Conjectural closed-form of $\int_0^1 \frac{\log^n (1-x) \log^{n-1} (1+x)}{1+x} dx$
Denote $f(k,j)=\int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^k(y)}{1-y} \, dy$. Then for $j, k>1$ (RHS denotes Beta derivatives)
$$U(k,j):=jf(k,j-1)+kf(j,k-1)=-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)$$
Which is direct by IBP, separation, Beta derivatives and reflection $y\to 1-y$: $$\small jf(k,j-1)= -(-\log (2))^{j+k}+ k \int_0^{\frac{1}{2}} \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$ $$\small =-(-\log (2))^{j+k}+ k \left(\int_0^{1}-\int_{\frac{1}{2}}^1 \right) \frac{\log ^j(1-y) \log ^{k-1}(y)}{y} \, dy$$ $$\small =-(-\log(2))^{j+k}+ k \left( \partial_a^{k-1} \partial_b^j B\right) (0,1)-kf(j,k-1)$$ Thus taking $\frac{\binom{n-1}{j-1} \binom{n}{k}}{\binom{n}{j} \binom{n-1}{k-1}}=\frac{j}{k}$ into account yields the important $\color{blue}{formula}$
$$\small \binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)=\frac{\binom{n}{k}\binom{n-1}{m-k} }{-k+m+1}U(k,m+1-k)$$
Now let $y\to\frac{1-x}{2}$ $$I_n=\int_0^{\frac{1}{2}} \frac{\log ^n(2 y) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$ Apply Binomial thm twice, extract $k=0$ $$I_n=\sum _{k=1}^n \sum _{j=0}^{n-1} \binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-j-k-1}(2)+\int_0^{\frac{1}{2}} \frac{\log ^n(2) \log ^{n-1}(2 (1-y))}{1-y} \, dy$$ Take Cauchy product $$I_n=\sum _{m=1}^{2n-1} \sum _{k+j=m}\binom{n}{k} \binom{n-1}{j} f(k,j) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Take care of range of $j,k$ $$\scriptsize I_n=\sum _{m=1}^n \sum _{k=1}^m \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+ \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \binom{n}{k} \binom{n-1}{m-k} f(k,m-k) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Let $k\to m+1-k$, take average $$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \left(\binom{n}{k} \binom{n-1}{m-k} f(k,m-k)+\binom{n}{m+1-k} \binom{n-1}{k-1} f(m+1-k,k-1)\right) \log ^{2n-m-1}(2)+\frac{\log ^{2 n}(2)}{n}$$ Use the $\color{blue}{formula}$ to simplify $$\scriptsize I_n=\frac{1}{2} \sum _{m=1}^n \sum _{k=1}^m \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1} U(k,m+1-k)+\frac{1}{2} \sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}U(k,m+1-k)+\frac{\log ^{2 n}(2)}{n}$$ Expand $U(k,m+1-k)$
$$ \scriptsize I_n=\frac{1}{2} \left(\sum _{m=1}^n \sum _{k=1}^m +\sum _{m=n+1}^{2 n-1} \sum _{k=m-n+1}^n\right) \frac{\binom{n}{k}\binom{n-1}{m-k} \log ^{-m+2 n-1}(2) }{-k+m+1}\left(k \underset{a\to 0}{\text{lim}}\underset{b\to 1}{\text{lim}}\frac{\partial ^{m}B(a,b)}{\partial a^{k-1}\, \partial b^{-k+m+1}}+(-1)^m \log ^{m+1}(2)\right)+\frac{\log ^{2 n}(2)}{n}$$
This is the final expression of $I_n$. According to Lemma $2.3$ in OP's article, all Beta derivatives in this expression lie in the algebra $\mathbb{Q}(\pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$, whence after adding up $\log(2)$ terms, $I_n$ lies in the extended $\mathbb{Q}(\log(2), \pi^2, \zeta(3), \zeta(5), \zeta(7), \cdots)$. QED.
Too long for a comment: By using the start I described in comments and then algebraic identities, I could reduce $\mathcal{I_4}$ to
$$\mathcal{I_4}=\log ^8(2)+\frac{31}{420} \log^2(2)\pi^6+4 \log (2) \underbrace{\int_0^1 \frac{\log ^3(1-t) \log ^3(t)}{t} \textrm{d}t}_{\text{Beta function}}+\log ^7(2)\int_0^{1/2} \frac{1}{1-t}\textrm{dt}\\+3 \log ^6(2)\int_0^{1/2}\frac{ \log (1-t)}{1-t}\textrm{d}t+4 \log ^6(2)\int_0^{1/2}\frac{ \log (t)}{1-t}\textrm{d}t+12 \log ^5(2) \int_0^{1/2}\frac{\log (1-t) \log (t)}{1-t}\textrm{d}t+12 \log ^3(2)\underbrace{\int_0^{1/2}\frac{ \log (1-t) \log ^3(t)}{1-t}\textrm{d}t}_{\text{Reducible to K}}+4 \log ^3(2)\int_0^{1/2} \frac{\log ^3(1-t) \log (t)}{1-t}\textrm{d}t\\+3 \log ^5(2)\int_0^{1/2}\frac{ \log ^2(1-t)}{1-t}\textrm{d}t+6 \log ^5(2)\int_0^{1/2}\frac{ \log ^2(t)}{1-t}\textrm{d}t+\frac{3}{5} \log ^2(2)\int_0^{1/2}\frac{ \log ^5(t)}{1-t}\textrm{d}t\\-\frac{3}{5} \log ^2(2)\int_0^{1/2}\frac{ \log ^5(1-t)}{1-t}\textrm{d}t+\log ^4(2) \int_0^{1/2} \frac{\log ^3(1-t)}{1-t}\textrm{d}t+4 \log ^4(2)\int_0^{1/2}\frac{ \log ^3(t)}{1-t}\textrm{d}t\\+\log ^3(2) \int_0^{1/2}\frac{\log ^4(t)}{1-t}\textrm{d}t+\underbrace{\int_0^{1/2}\frac{\log ^3(1-t) \log ^4(t)}{1-t}\textrm{d}t}_{\text{Reducible to $J_3$}}+18 \log ^4(2) \underbrace{\int_0^{1/2}\frac{ \log (1-t) \log ^2(t)}{1-t}\textrm{d}t}_{\textrm{Reducible to $J_1$}}+12 \log ^4(2)\int_0^{1/2}\frac{ \log ^2(1-t) \log (t)}{1-t}\textrm{d}t+3 \log ^2(2)\int_0^{1/2}\frac{ \log ^4(1-t) \log (t)}{1-t}\textrm{d}t\\+18 \log ^3(2)\underbrace{\int_0^{1/2}\frac{ \log ^2(1-t) \log ^2(t)}{1-t}\textrm{d}t}_{\text{Reducible to $K$}}+18 \log ^2(2) \underbrace{\int_0^{1/2}\frac{\log ^2(1-t) \log ^3(t)}{1-t} \textrm{d}t}_ {\text{Reducible to $J_2$}}.$$
I considered the auxiliary results
$$J_n=\int_0^{1/2} \frac{\log^n(1-x)\log^{n+1}(x)}{1-x}\textrm{d}x=-\frac{1}{2(1+n)}\log^{2(n+1)}(2)+\frac{1}{2}\lim_{\substack{x\to0 \\ y \to 1}}\frac{\partial^{2n+1}}{\partial x^n \partial y^{n+1}}\operatorname{B}(x,y)$$ and $$ K=\int_{0}^{1/2} \frac{\log^2(x)\log^2(1-x)}{x}\textrm{d}x$$ $$=\frac{1}{8}\zeta(5)-2\zeta(2)\zeta(3)-\frac{2}{3}\log^3(2)\zeta(2)+\frac{7}{4}\log^2(2)\zeta(3)-\frac{1}{15}\log^5(2) $$ $$+4\log(2)\operatorname{Li}_4\left(\frac{1}{2}\right)+4\operatorname{Li}_5\left(\frac{1}{2}\right),$$ which are both calculated in the book (Almost) Impossible Integrals, Sums, and Series.
A short note: For the generalization the key is to figure out which groups of integrals to take together for transformations after the first step I described in comments, where to further use algebraic identities to get those expected magical cancellations like in the case above. The rest is trivial. Also, I skipped giving references for the trivial integrals above.
My approach to $I_3$:
Starting with the algebraic identity $20a^3b^2=(a+b)^5+(a-b)^5-2a^5-10ab^4$ we can write
$$20\int_0^1\frac{\ln^3(1-x)\ln^2(1+x)}{1+x}\ dx\\=\int_0^1\frac{\ln^5(1-x^2)}{1+x}+\int_0^1\frac{\ln^5\left(\frac{1-x}{1+x}\right)}{1+x}-2\int_0^1\frac{\ln^5(1-x)}{1+x}-10\int_0^1\frac{\ln(1-x)\ln^4(1+x)}{1+x}\ dx$$
The first integral can be calculated the same way Cornel did here
$$\int_0^1\frac{\ln^5(1-x^2)}{1+x}dx=\int_0^1(1-x)\frac{\ln^5(1-x^2)}{1-x^2}dx\overset{x^2=y}{=}\frac12\int_0^1\frac{1-\sqrt{y}}{\sqrt{y}}.\frac{\ln^5(1-y)}{1-y}dy$$ $$\overset{IBP}{=}-\frac1{24}\int_0^1\frac{\ln^6(1-y)}{y^{3/2}}dy=-\frac{1}{24}\lim_{x\mapsto-1/2\\y\mapsto1}\frac{\partial^6}{\partial y^6}\text{B}(x,y)$$
$$\boxed{=\frac{16}3\ln^62-40\ln^42\zeta(2)+160\ln^32\zeta(3)-270\ln^22\zeta(4)+720\ln2\zeta(5)\\-240\ln2\zeta(2)\zeta(3)-\frac{1185}{4}\zeta(6)+120\zeta^2(3)}$$
The second integral can be simplified via subbing $\frac{1-x}{1+x}=y$:
$$\int_0^1\frac{\ln^5\left(\frac{1-x}{1+x}\right)}{1+x}\ dx=\int_0^1\frac{\ln^5 y}{1+y}\ dy\\=-\sum_{n=1}^\infty (-1)^n\int_0^1 y^{n-1}\ln^5 y\ dy=5!\sum_{n=1}^\infty\frac{(-1)^n}{n^6}=\boxed{-\frac{465}{4}\zeta(6)}$$
and lets set $1-x=y$ for the third integral:
$$\int_0^1\frac{\ln^5(1-x)}{1+x}\ dx=\int_0^1\frac{\ln^5y}{2-y}\ dy\\=\sum_{n=1}^\infty\frac1{2^n}\int_0^1 y^{n-1}\ln^5 y\ dy=-5!\sum_{n=1}^\infty\frac{1}{2^n n^6}=\boxed{-120\operatorname{Li}_6(1/2)}$$
For the last integral, we set $1+x=y$
$$\int_0^1\frac{\ln(1-x)\ln^4(1+x)}{1+x}\ dx=\int_1^2\frac{\ln(2-y)\ln^4y}{y}\ dy\\=\ln2\int_1^2\frac{\ln^4y}{y}\ dx+\int_1^2\frac{\ln(1-y/2)\ln^4y}{y}\ dy\\=\frac15\ln^62-\sum_{n=1}^\infty\frac{1}{n2^n}\int_1^2 y^{n-1}\ln^4y\ dy\\=\frac15\ln^62-\sum_{n=1}^\infty\frac1{n2^n}\left(24\frac{2^n}{n^5}-24\frac{2^n\ln2}{n^4}+12\frac{2^n\ln^22}{n^3}-4\frac{2^n\ln^32}{n^2}+\frac{2^n\ln^42}{n}-\frac{24}{n^5}\right)\\\boxed{=\frac15\ln^62-24\zeta(6)+24\ln2\zeta(5)-12\ln^22\zeta(4)+4\ln^32\zeta(3)-\ln^42\zeta(2)+24\operatorname{Li}_6(1/2)}$$
By combining the boxed results, the closed form of $I_3$ follows: