I managed to find
$$S=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}=20\beta(6)+\frac12\zeta(2)\beta(4)-\frac{7\pi^3}{32}\zeta(3)-\frac{31\pi}{8}\zeta(5)$$
where $\beta(a)$ is the Dirichlet Beta function.
and would like to see different approaches.
Here is my work
Following the same approach here
\begin{align} 6S&=6\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}\\ &=\sum_{n=1}^\infty(-1)^{n-1}H_n^{(2)}\int_0^1-x^{2n}\ln^3x\ dx\\ &=\int_0^1\ln^3x\sum_{n=1}^\infty(-x^2)^nH_n^{(2)}\ dx\\ &=\int_0^1\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\ &=\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{x\mapsto 1/x}\\ &=\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln^3x\operatorname{Li}_2(-1/x^2)}{1+x^2}\ dx\\ &\left\{\small{\text{add the integral} \int_0^1\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx=6S\text{ to both sides}}\right\}\\ 12S&=\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\int_0^1\frac{\ln^3x[\color{red}{\operatorname{Li}_2(-x^2)+\operatorname{Li}_2(-1/x^2)}]}{1+x^2}\ dx\\ S&=\frac1{12}\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx+\frac1{12}\int_0^1\frac{\ln^3x[\color{red}{-2\ln^2x-\zeta(2)}]}{1+x^2}\ dx\\ &=\frac1{12}\underbrace{\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx}_{I}-\frac16\underbrace{\int_0^1\frac{\ln^5x}{1+x^2}\ dx}_{-120\beta(6)}-\frac{\zeta(2)}{12}\underbrace{\int_0^1\frac{\ln^3x}{1+x^2}\ dx}_{-6\beta(4)}\\ &=\frac1{12}I+20\beta(6)+\frac12\zeta(2)\beta(4)\tag1 \end{align}
\begin{align} I&=\int_0^\infty\frac{\ln^3x\operatorname{Li}_2(-x^2)}{1+x^2}\ dx\\ &=\int_0^\infty\frac{\ln^3x}{1+x^2}\left(\int_0^1\frac{x^2\ln y}{1+yx^2}\ dy\right)\ dx\\ &=\int_0^1\ln y\left(\int_0^\infty\frac{x^2\ln^3x}{(1+x^2)(1+yx^2)}\ dx\right)\ dy\\ &=\int_0^1\frac{\ln y}{1-y}\left(\int_0^\infty\frac{\ln^3x}{1+yx^2}\ dx-\underbrace{\int_0^\infty\frac{\ln^3x}{1+x^2}\ dx}_{0}\right)\ dy\\ &=\int_0^1\frac{\ln y}{1-y}\left(-\frac{\pi^3}{16}.\frac{\ln y}{\sqrt{y}}-\frac{\pi}{16}.\frac{\ln^3y}{\sqrt{y}}\right)\ dy,\quad \sqrt{y}=x\\ &=-\frac32\pi^3\underbrace{\int_0^1\frac{\ln^2x}{1-x^2}\ dx}_{\frac74\zeta(3)}-2\pi\underbrace{\int_0^1\frac{\ln^4x}{1-x^2}\ dx}_{\frac{93}{4}\zeta(5)}\\ &=-\frac{21}{8}\pi^3\zeta(3)-\frac{93}{2}\pi\zeta(5)\tag2 \end{align}
Plug $(2)$ and $(1)$ we get
$$S=20\beta(6)+\frac12\zeta(2)\beta(4)-\frac{7}{32}\pi^3\zeta(3)-\frac{31}{8}\pi\zeta(5)$$
Thank to @Zacky for spotting couple mistakes in my solution