Using the fact that
$$\Re\left\{\frac{\ln(1-ix)}{1-ix}\right\}=\frac{\arctan(x)}{x}-\frac{\arctan(x)}{x(1+x^2)}+\frac{\ln(1+x^2)}{2(1+x^2)}$$
we have
$$\Re\left\{\int_0^1\frac{\ln^2x\ln(1-ix)}{1-ix}dx\right\}$$
$$=\int_0^1\frac{\ln^2x\arctan(x)}{x}dx-\int_0^1\frac{\ln^2x\arctan(x)}{x(1+x^2)}dx+\frac12\int_0^1\frac{\ln^2x\ln(1+x^2)}{1+x^2}$$
$$=I_1-I_2+\frac12I_3$$
$$I_1=\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}\int_0^1 x^{2n}\ln^2xdx=2\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^4}=2\beta(4)$$
$I_2$ was evaluated by a friend (Kartick Betal).
$$I_2=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\underbrace{\int_1^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx}_{\displaystyle x\mapsto 1/x}$$
$$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\int_0^1 \frac{x\ln^2x\left(\frac{\pi}{2}-\arctan x\right)}{1+x^2}\ dx$$
$$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\int_0^1 \frac{x\ln^2x}{1+x^2}\ dx+\int_0^1 \frac{x\ln^2x\arctan x}{1+x^2}\ dx$$
$$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{\pi}{2}\cdot\frac3{16}\zeta(3)+\int_0^1 \left(\frac1x-\frac1{x(1+x^2)}\right)\ln^2x\arctan xdx$$
$$=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+\int_0^1 \frac{\ln^2x\arctan x}{x}\ dx-I$$
$$\Longrightarrow 2I_2=\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx-\frac{3\pi}{32}\zeta(3)+2\beta(4)$$
For the remaining integral, write $\arctan x=\int_0^1\frac{x}{1+x^2y^2}\ dy\ $, we get
$$\int_0^\infty \frac{\ln^2x\arctan x}{x(1+x^2)}\ dx=\int_0^\infty \frac{\ln^2x}{x(1+x^2)}\left(\int_0^1\frac{x}{1+x^2y^2}\ dy\right)\ dx$$
$$=\int_0^1\frac{1}{1-y^2}\left(\int_0^\infty\frac{\ln^2x}{1+x^2}\ dx-\int_0^\infty\frac{y^2\ln^2x}{1+x^2y^2}\ dx\right)\ dy$$
$$=\int_0^1\frac{1}{1-y^2}\left(\frac{\pi^3}{8}-\frac{y\pi^3}{8}-\frac{y\pi\ln^2y}{2}\right)\ dy$$
$$=\frac{\pi^3}{8}\int_0^1\frac{1-y}{1-y^2}\ dy-\frac{\pi}2\int_0^1\frac{y\ln^2y}{1-y^2}\ dy$$
$$=\frac{\pi^3}{8}\int_0^1\frac{1}{1+y}\ dy-\frac{\pi}{16}\int_0^1\frac{\ln^2y}{1-y}\ dy$$
$$=\frac{\pi^3}{8}\ln2-\frac{\pi}{8}\zeta(3)$$
Plug in this result, we get
$$I_2=\frac{\pi^3}{16}\ln(2)-\frac{7\pi}{32}\zeta(3)+\beta(4)$$
$$I_3=\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx-\underbrace{\int_1^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x\mapsto1/x}$$
$$=\underbrace{\int_0^\infty\frac{\ln^2x\ln(1+x^2)}{1+x^2}\ dx}_{\large x^2\mapsto x}-I_3+2\int_0^1\frac{\ln^3x}{1+x^2}\ dx$$
$$\Longrightarrow 2I_3=\frac18\int_0^\infty\frac{\ln^2x\ln(1+x)}{\sqrt{x}(1+x)}\ dx+2(-6\beta(4))$$
$$I_3=\frac1{16}\lim_{a\ \mapsto1/2\\b\ \mapsto1/2}\frac{-\partial^3}{\partial a^2\partial b}\text{B}(a,b)-6\beta(4)$$
$$=\frac{7\pi}{8}\zeta(3)+\frac{\pi^3}{8}\ln(2)-6\beta(4)$$
For the LHS integral, write $\frac{\ln(1-ix)}{1-ix}=-\sum_{n=1}^\infty (ix)^{n-1}H_{n-1}$
$$\Re\left\{\int_0^1\frac{\ln^2x\ln(1-ix)}{1-ix}dx\right\}=\Re\left\{-\sum_{n=1}^\infty i^{n-1}H_{n-1}\int_0^1 x^{n-1}\ln^2xdx\right\}$$
$$=\Im\left\{2\sum_{n=1}^\infty \frac{i^{n}H_{n-1}}{n^3}\right\}=\Im\left\{2\sum_{n=1}^\infty \frac{i^{n}H_{n}}{n^3}-2\sum_{n=1}^\infty \frac{i^{n}}{n^4}\right\}$$
use $\Im \left\{\sum_{n=1}^\infty i^n f(n)\right\}=\sum_{n=0}^\infty (-1)^n f(2n+1)$
$$=2\sum_{n=0}^\infty \frac{(-1)^nH_{2n+1}}{(2n+1)^3}-2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^4}$$
$$=2\sum_{n=0}^\infty \frac{(-1)^nH_{2n+1}}{(2n+1)^3}-2\beta(4)$$
Collect all results and use $\beta(4)=\frac1{768}\left(\psi^{(3)}\left(\frac14\right)-8\pi^4\right)$ we find
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$