A conjectured result for $\sum_{n=1}^\infty\frac{(-1)^n\,H_{n/5}}n$
Here is how one can compute $\mathcal S_m$ for arbitrary $m$.
Your formula (3) can be rewritten as $$\mathcal{S}_m=-m\int_0^1\frac{\ln\frac{1+z}{2}}{z^m-1}z^{m-1}dz=-m\sum_{k=0}^{m-1}\alpha_{km}\int_0^1\frac{\ln\frac{1+z}{2}}{z-e^{2\pi i k/m}}dz,$$ where $$\alpha_{km}=\lim_{\quad z\to\; \exp{\frac{2\pi i k}m}}\frac{z^{m-1}\left(z-e^{2\pi i k/m}\right)}{z^m-1}=\frac{e^{2\pi i k(m-1)/m}}{\prod_{n\ne k}\left(e^{2\pi i k/m}-e^{2\pi i n/m}\right)}=\frac1m.$$ Note in particular that the last expression is independent of $k$.
The remaining integrals can be computed in terms of polylogarithms: $$I\left(\zeta\right)=\int_0^1\frac{\ln\frac{1+z}{2}}{z-\zeta}dz= \operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)- \operatorname{Li}_2\left(\frac{1}{1+\zeta}\right)+ \ln2\ln\frac{\zeta}{1+\zeta}.\tag{1}$$ We have in particular $$I\left(1\right)=\frac{\pi^2}{12}-\frac{\ln^22}{2},\qquad I\left(-1\right)=-\frac{\ln^22}{2}. $$
This implies that $$\mathcal{S}_m=-\sum_{k=0}^{m-1}I\left(e^{2\pi i k/m}\right), \tag{2}$$ with $I\left(\zeta\right)$ defined by (1). It is clear that under the sum the elementary pieces of $I(\zeta)$ simplify. It might happen that something nice happens also with the dilogarithmic ones.
Update 1 (how to simplify one of the two sums of dilogarithms to an elementary expression):
- This formula for $\operatorname{Li}_2\left(e^{2\pi i \mathbb{Q}}\right)$ implies that $$\sum_{k=0}^{m-1}\operatorname{Li}_2\left(e^{2\pi i k/m}\right)=\frac{\pi^{2}}{6m},\qquad \sum_{k=0}^{m-1}\operatorname{Li}_2\left(-e^{2\pi i k/m}\right)=\begin{cases}\quad \frac{\pi^{2}}{6m},\quad & m\text{ even}, \\ -\frac{\pi^{2}}{12m},\quad & m \text{ odd}. \end{cases} $$
We also have the identity $\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2\left(1-z\right)$, which can be rewritten as $$\operatorname{Li}_2\left(\frac{1}{\zeta+1}\right)=-\operatorname{Li}_2\left(-\zeta^{-1}\right)-\frac12\ln^2\frac{\zeta}{\zeta+1}.$$
Combining both results, we obtain for odd $m$ $$\sum_{k=0}^{m-1}\operatorname{Li}_2\left(\frac{1}{1+e^{2\pi i k/m}}\right)= \frac{\pi^2}{12m}-\frac12\sum_{k=0}^{m-1}\ln^2\left(1+e^{-2\pi i k /m}\right),$$ and for even $m$ $$\sum_{k=0 | k\neq \frac{m}{2}}^{m-1}\operatorname{Li}_2\left(\frac{1}{1+e^{2\pi i k/m}}\right)= \frac{\pi^2(m-1)}{6m}-\frac12\sum_{k=0| k\neq \frac{m}{2}}^{m-1}\ln^2\left(1+e^{-2\pi i k /m}\right).$$
Update 2 (simplification of the remaining sum for even $m$):
We can use again the identity $\operatorname{Li}_2\left(z\right)=-\operatorname{Li}_2\left(\frac{z}{z-1}\right)-\frac12\ln^2\left(1-z\right)$ to show that $$\operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)+\operatorname{Li}_2\left(\frac{2}{1-\zeta}\right)=-\frac12\ln^2\frac{\zeta-1}{\zeta+1}.$$ For even $m$, if $\zeta$ is an $m$th root of unity, then so is $-\zeta$, which simplifies the second sum to an elementary expression: $$\sum_{k=0 | k\neq \frac{m}{2}}^{m-1}\operatorname{Li}_2\left(\frac{2}{1+e^{2\pi i k/m}}\right)=\frac{\pi^2}{12}-\frac{\ln^2 2}{2}-\frac12\sum_{k=1}^{\frac{m}{2}-1}\ln^2\frac{e^{2\pi i k/m}-1}{e^{2\pi i k/m}+1}.$$ Altogether, this leads to evaluation \begin{align*} \mathcal{S}_{2n}&=\frac{\ln 2\ln 8n^2}{2}-\frac{\pi^2}{12n}+\frac12 \sum_{k=1}^{n-1}\ln^2\frac{e^{\pi i k/n}-1}{e^{\pi i k/n}+1}-\frac12\sum_{k=0| k\neq n}^{2n-1}\ln^2\left(1+e^{-\pi i k /n}\right)=\\ &=\ln 2\ln 2n-\frac{\pi^2\left(n^2+1\right)}{24n}-\sum_{k=1}^{n-1}\ln\left(2\sin\frac{\pi k}{2n}\right)\ln\left(2\cos\frac{\pi k}{2n}\right). \tag{$\spadesuit$} \end{align*}
Update 3 (partial simplification for odd $m$)
Let us denote $m=2n+1$. We will use the identity $$\operatorname{Li}_2\left(\frac{2}{1+\zeta}\right)=\operatorname{Li}_2\left(\frac{\zeta+1}{\zeta-1}\right)+\frac12 \ln^2 \frac{\zeta-1}{\zeta+1} -\ln\left(-\frac{2}{1+\zeta}\right)\ln\frac{\zeta-1}{\zeta+1}-\frac{\pi^2}{6}. \tag{3}$$ Now the key three facts are that
If $\zeta$ is $\zeta^m=1$, then so is $\zeta^{-1}$.
Under replacement $\zeta\leftrightarrow \zeta^{-1}$, the dilogarithm argument on the right of (3) changes its sign.
There is an identity $\operatorname{Li}_2\left( z\right)+ \operatorname{Li}_2\left( - z\right)=\frac{1}{2}\operatorname{Li}_2\left( z^2\right)$.
Put altogether, this leads to \begin{align*}\sum_{k=0}^{m-1}\operatorname{Li}_2\left(\frac{2}{1+e^{\frac{2\pi i k}{m}}}\right)=&\frac12\sum_{k=1}^n\operatorname{Li}_2\left(-\cot^2\frac{\pi k}{2n+1}\right)+\frac{4n^2+3n+2}{2n+1}\cdot\frac{\pi^2}{12}+\\ &+\sum_{k=1}^n\ln\left(\tan\frac{\pi k}{2n+1}\right) \ln\left(\frac12\sin\frac{2\pi k}{2n+1}\right). \end{align*} Combining this with the previous results of Update 1, we finally arrive at \begin{align} \nonumber\mathcal{S}_{2n+1}=& -\frac12\sum_{k=1}^n\operatorname{Li}_2\left(-\cot^2\frac{\pi k}{2n+1}\right)-\sum_{k=1}^n\ln^2\sin\frac{\pi k}{2n+1} +\\ &+\left(n+\frac12\right)\ln^2 2-\frac{\left(2n^2+n+1\right)\pi^2}{12\left(2n+1\right)}.\tag{$\clubsuit$} \end{align} Remark. For $n=2$ (i.e. $m=5$) this sum contains two dilogarithms $\operatorname{Li}_2\left(-1\pm\frac{2}{\sqrt5}\right)$. One should be able to reduce them to just one $\operatorname{Li}_2\left(\frac15\right)$ using a suitable identity.
This is not a novel result, but rather an alternate derivation of @Start wearing purple's result. I tried to keep everything simple and explained, which resulted in a slightly verbose solution. So if you are not interested in details, you may follow only the tagged equations until Step 3.
Step 1 (Reduction of Integral). We begin with the following formula:
$$ \mathcal{S}_m = \int_{0}^{1} \frac{\log(1-x^m)}{1+x} \, dx = \sum_{\omega \ : \ \omega^m = 1} \int_{0}^{1} \frac{\log(1 - \omega x)}{1+x} \, dx. \tag{1} $$
In order to work with the RHS, we consider it as a function of $\omega \in \Bbb{C} \setminus (1, \infty)$. Using the differentiation under the integral sign technique, we find that
\begin{align*} \int_{0}^{1} \frac{\log(1 - \omega x)}{1+x} \, dx &= \int_{0}^{\omega} \left( \frac{d}{dz} \int_{0}^{1} \frac{\log(1 - zx)}{1+x} \, dx \right) \, dz\\ &= \int_{0}^{\omega} \left( \int_{0}^{1} \left( \frac{1}{1+x} - \frac{1}{1-zx}\right) dx \right) \frac{dz}{1+z} \\ &= \int_{0}^{\omega} \left( \frac{\log(1-z)}{z} - \frac{\log\left(\frac{1-z}{2}\right)}{1+z} \right) \, dz \\ &= -\operatorname{Li}_2(\omega) + I(\omega), \end{align*}
where the integral w.r.t. $z$ is taken inside the region $\Bbb{C}\setminus(1, \infty)$ and the function $I(\omega)$ is defined on $\Bbb{C}\setminus(1,\infty)$ as
$$ I(\omega) := -\int_{0}^{\omega} \frac{\log\big(\frac{1-z}{2}\big)}{1+z} \, dz. \tag{*} $$
Step 2 (Properties of $I(\omega)$). Why we consider this integral is that it satisfies the following two properties:
Using the identity $\text{(1)}$ and the relation we have developed, $\mathcal{S}_m$ is written as $$ \mathcal{S}_m =\sum_{\omega \ : \ \omega^m = 1} (I(\omega) - \operatorname{Li}_2(\omega)) = -\frac{\zeta(2)}{m} + \sum_{\omega \ : \ \omega^m = 1} I(\omega). \tag{2} $$ This follows from the multiplication theorem for polylogarithm.
More importantly, for $\omega \in \Bbb{C}$ outside $(-\infty, 1) \cup (1, \infty)$, we have $$ I(-\omega) + I(\omega) = \log^2 2 - \log\big(\tfrac{1+\omega}{2}\big)\log\big(\tfrac{1-\omega}{2}\big). \tag{3} $$ Here, we take a convention that the log part is $0$ when $\omega = \pm 1$. This is consistent both with the limit as $\omega \to 1$ or $\omega \to -1$ and with the actual known values of $I(1)+I(-1)$. Notice also that $\text{(3)}$ easily follows from integration by parts: $$ I(-\omega) = \left[ -\log\big(\tfrac{1+z}{2}\big)\log\big(\tfrac{1-z}{2}\big) \right]_{0}^{-\omega} - \int_{0}^{-\omega} \frac{\log\big(\frac{1+z}{2}\big)}{1-z} \, dz. $$
Step 3 (Formula for $\mathcal{S}_{2p}$). When $m = 2p$, as @Start wearing purple observed, $\omega^m = 1$ implies $(-\omega)^m = 1$. Thus combining $\text{(2)}$ and $\text{(3)}$, we have
\begin{align*} \mathcal{S}_{2p} &= -\frac{\pi^2}{12p} + \frac{1}{2} \sum_{\omega \ : \ \omega^{2p} = 1} ( I(\omega) + I(-\omega)) \\ &= -\frac{\pi^2}{12p} + \frac{1}{2} \sum_{\omega \ : \ \omega^{2p} = 1} \left( \log^2 2 - \log\left(\frac{1+\omega}{2}\right)\log\left(\frac{1-\omega}{2}\right) \right). \end{align*}
(Here, as pointed out in Step 2, we consider $\log(\frac{1+\omega}{2})\log(\frac{1-\omega}{2}) = 0$ when $\omega = \pm 1$.) Finally, we can simplify the last summation by considering only $\omega$ with $\Im(\omega) > 0$ as follows:
\begin{align*} \mathcal{S}_{2p} &= p\log^2 2 - \frac{\pi^2}{12p} - \Re \sum_{\substack{\omega \ : \ \omega^{2p} = 1 \\ \Im(\omega) > 0}} \log\left(\frac{1+\omega}{2}\right)\log\left(\frac{1-\omega}{2}\right) \\ &= p\log^2 2 - \frac{\pi^2}{12p} - \Re \sum_{k=1}^{p-1} \log\left(\frac{1+e^{ik\pi/p}}{2}\right)\log\left(\frac{1+e^{i(k-p)\pi/p}}{2}\right) \\ &= p\log^2 2 - \frac{\pi^2}{12p} - \sum_{k=1}^{p-1} \left( \log\left(\cos\frac{\pi k}{2p}\right)\log\left(\sin\frac{\pi k}{2p}\right) + \frac{\pi^2}{4p^2}k(p-k) \right) \\ &= p\log^2 2 - \frac{(p^2+1)\pi^2}{24p} - \sum_{k=1}^{p-1} \log\left(\cos\frac{\pi k}{2p}\right)\log\left(\sin\frac{\pi k}{2p}\right). \end{align*}
This solution relies on the symmetry between $I(\omega)$ and $I(-\omega)$, so I doubt that this will work for odd $m$.
And what about this one?
$$\mathcal{S}_6 \stackrel{?}{=} \ln^2(2) + \ln(2)\ln(3) -\frac{5\pi^2}{36} .$$
Furthermore, you probably know that
$$ \operatorname{Li}_2\left(\frac{\sqrt5 - 1}{2}\right) = \frac{\pi^2}{10} - \ln^2\left(\varphi\right) = \frac{\pi^2}{10} - \ln^2(2) -\ln^2\left(1+\sqrt{5}\right) + 2\ln(2)\ln\left(1+\sqrt{5}\right), $$
where $\varphi = \tfrac{1}{2}\left(1+\sqrt5\right)$ is the golden ratio.
At last, it seems to me there are always some $$\operatorname{Li}_2\left({\tfrac12}\right) = \frac{\pi^2}{12} - \frac{1}{2}\ln^2(2)$$ and $$\operatorname{Li}^2_1\left({\tfrac12}\right) = \ln^2(2)$$ behind the scenes.
I've found some relations between particular $\mathcal{S}_m$ values. For example: $$\begin{align} \mathcal{S}_{1} &=\tfrac{9}{4}\mathcal{S}_{2}-\mathcal{S}_4\\ \mathcal{S}_{1} &=\tfrac{9}{5}\mathcal{S}_{1/3}-\tfrac{8}{5}\mathcal{S}_{1/2}\\ \mathcal{S}_{1} &= \tfrac{1}{2}\mathcal{S}_{1/2}+\tfrac{3}{8}\mathcal{S}_2\\ \mathcal{S}_{1} &=\tfrac{3}{7}\mathcal{S}_{1/3}+\tfrac{2}{7}\mathcal{S}_2\\ \mathcal{S}_{1} &=\tfrac{3}{5}\mathcal{S}_{1/2}+\tfrac{1}{5}\mathcal{S}_4\\ \mathcal{S}_{1} &=\tfrac{27}{55}\mathcal{S}_{1/3}+\tfrac{8}{55}\mathcal{S}_4\\ \mathcal{S}_{1} &=\mathcal{S}_{1/2}+\mathcal{S}_{4} -\tfrac{3}{2}\mathcal{S}_{2}\\ \mathcal{S}_{1} &=\tfrac{1}{10}\mathcal{S}_{4}+\tfrac{9}{10}\mathcal{S}_{1/3} - \tfrac{1}{2}\mathcal{S}_{1/2}\\ \end{align}$$ or my favorite one: $$ 3\,\mathcal{S}_{4}+5\,\mathcal{S}_{1/2} = 3\,\mathcal{S}_{1/3}+5\,\mathcal{S}_{2}. $$