How to calculate $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}h_n}{n^3}$ where $h_n=\sum_{k=1}^{n}\frac{1}{2k-1}$?

Noting that $$\sum_{k = 1}^n \frac{1}{2k - 1} = H_{2n} - \frac{1}{2} H_n,$$ where $H_n$ denotes the $n$th harmonic number, your sum $S$ can be re-written in terms of the following two Euler sums: $$S = \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} \left (H_{2n} - \frac{1}{2} H_n \right ) = -\sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^3} + \frac{1}{2} \frac{(-1)^n H_n}{n^3}.$$

Dealing with these two Euler sums, their values can be found from the following generating function \begin{align} \sum^\infty_{n=1}\frac{H_n}{n^3}z^n &=2{\rm Li}_4(x)+{\rm Li}_4\left(\frac{x}{x-1}\right)-{\rm Li}_4(1-x)-{\rm Li}_3(x)\ln(1-z)-\frac{1}{2}{\rm Li}_2^2\left(\frac{x}{x-1}\right)\\ &+\frac{1}{2}{\rm Li}_2(x)\ln^2(1-x)+\frac{1}{2}{\rm Li}_2^2(x)+\frac{1}{6}\ln^4(1-x)-\frac{1}{6}\ln{x}\ln^3(1-x)\\ &+\frac{\pi^2}{12}\ln^2(1-x)+\zeta(3)\ln(1-x)+\frac{\pi^4}{90},\tag1 \end{align} which is proved in this answer here.

Setting $x = -1$ in (1) gives \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{n^3}=2{\rm Li}_4\left(\tfrac{1}{2}\right)-\frac{11\pi^4}{360}+\frac{7}{4}\zeta(3)\ln{2}-\frac{\pi^2}{12}\ln^2{2}+\frac{1}{12}\ln^4{2}, \end{align} while setting $x = i$ in (1) gives \begin{align} \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^3} &= 8 \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{(2n)^3}\\ &= 8 \operatorname{Re} \sum_{n = 1}^\infty \frac{H_n}{n^3} i^n\\ &= -16 \operatorname{Re} \operatorname{Li}_4(1 + i) + \frac{29 \pi^4}{288} + \frac{35}{8} \zeta (3) \ln 2 + \frac{\pi^2}{8} \ln^2 2. \end{align}

Substituting these two values for the Euler sums back into the expression for the sum $S$ gives $$S = 16 \operatorname{Re} \operatorname{Li}_4 (1 + i) - \frac{167}{1440} \pi^4 - \frac{7}{2} \zeta (3) \ln 2 - \frac{\pi^2}{6} \ln^2 2 + \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{1}{24} \ln^4 2.\tag2$$

The above result can be further simplified since recently a closed form for $\operatorname{Re} \operatorname{Li}_4 (1 + i)$ has been found. As (for a proof of this, see here):

$$\operatorname{Re} \operatorname{Li}_4 (1 + i) = -\frac{5}{16} \operatorname{Li}_4 \left (\frac{1}{2} \right ) + \frac{97 \pi^4}{9216} + \frac{\pi^2}{48} \ln^2 2 - \frac{5}{384} \ln^4 2,$$

the value for the required sum becomes

$$\sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} \left (H_{2n} - \frac{1}{2} H_n \right ) = -4 \operatorname{Li}_4 \left (\frac{1}{2} \right ) - \frac{7}{2} \zeta (3) \ln 2 + \frac{151 \pi^4}{2880} - \frac{1}{6} \ln^4 2 + \frac{\pi^2}{6} \ln^2 2.$$