Considering the algebraic identity
\begin{align*}
&(a-b)^3b = a^3b - 3a^2b^2 + 3ab^3 - b^4 = -2a^3b +3(a^3b+ab^3) -3a^2b^2 -b^4\\
&\Longrightarrow \ \ \ 2a^3b = -{b^4 \over 2} -{b^4 + 6a^2b^2\over 2} + 3(a^3b+ab^3) - (a-b)^3b
\end{align*} with $a = \ln(1-x)$ and $b= \ln (1+x)$ it follows that
\begin{align*}
2\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =& - \frac 1 2\int_0^1 {\ln^4(1+x)\over x}d x \\
&-\frac 12 \int_0^1 \frac{\ln^4(1+x) + 6\ln^2(1-x)\ln^2(1+x)}{x}dx\\
&+3\int_0^1 \frac{\ln^3(1-x)\ln(1+x) + \ln(1-x)\ln^3(1+x)}{x}dx\\
&- \int_0^1 \frac{\ln^3\left(\frac{1-x}{1+x}\right)\ln(1+x)}{x}dx\\
=:& -I_1 - I_2 + I_3 -I_4.
\end{align*}
For $I_1$, make substitution $y = \frac x {1+x}$ to get:
\begin{align*}
I_1 =& \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{y(1-y)} dy \\
=& \frac 1 2\underbrace{ \int_0^{\frac 12} \frac{\ln^4(1-y)}{y} dy}_{z=1-y}+ \frac 1 2 \int_0^{\frac 12} \frac{\ln^4(1-y)}{1-y} dy\\
=& \frac 1 2 \int_{\frac 1 2 }^1 \frac{\ln^4 z} {1-z} dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \int_{\frac 1 2}^1 z^{n-1}\ln^4 z\ dz + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \frac{\partial^4}{\partial n^4}\left[\frac 1 n - \frac 1 {n2^n}\right] + \frac {\ln^5 2}{10}\\
=& \frac 12 \sum_{n=1}^\infty \left[\frac{24}{n^5} - \frac {24}{n^52^n} - \frac{24 \ln 2}{n^42^n}-\frac{12\ln^2 2}{n^3 2^n}-\frac{4\ln^3 2}{n^2 2^n} - \frac{\ln^4 2}{n2^n}\right] + \frac {\ln^5 2}{10}\\
=&12\zeta(5) - 12\text{Li}_5(1/2) - 12\ln 2 \text{Li}_4(1/2) -6\ln^2 2 \text{Li}_3(1/2) -2\ln^3 2\text{Li}_2(1/2)-\frac {2}{5}\ln^5 2\\
=&\boxed{-12\Big(\text{Li}_5(1/2) + \ln 2\text{Li}_4(1/2)-\zeta(5)\Big)-{21 \over 4}\zeta(3)\ln^2 2 +{1\over 3} \pi^2 \ln^3 2-{2 \over 5} \ln^5 2}
\end{align*} where the well-known values
\begin{align*}\text{Li}_2(1/2) = {\pi^2 \over 12}-{\ln^2 2\over 2} , \qquad \text{Li}_3(1/2) ={7\zeta(3) \over 8} -{\pi^2 \ln 2\over 12} + {\ln^3 2 \over 6}
\end{align*} are used.
Actually, $I_2$ was already evaluated by the OP here using the algebraic identity $$b^4 + 6a^2b^2 = \frac {(a-b)^4} 2+\frac{(a+b)^4}{2} -a^4.$$
It holds that
$$
\boxed{I_2 = \frac {21}{8} \zeta(5).}
$$
In fact, the value of $I_3$ can also be found in the previous answer of @Przemo's. For $I_3$, one can use the algebraic relation $3(a^3b + ab^3) =\frac 3 8 \left[ (a+b)^4 - (a-b)^4\right]$.
This gives
\begin{align*}
I_3=& \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4(1-x^2)}{x} dx}_{x^2 = y} - \underbrace{\frac 3 8 \int_0^1 \frac{\ln^4\left(\frac{1-x}{1+x}\right)}{x} dx}_{\frac{1-x}{1+x} = y}\\
=&\frac 3 {16}\underbrace{\int_0^1 \frac{\ln^4(1-y)}{y} dy }_{1-y\mapsto y}- \frac 3 4 \int_0^1 \frac{\ln^4 y}{1-y^2} dy\\
=&\frac 3 {16}\int_0^1 \frac{\ln^4 y}{1-y} dy - \frac 3 4 \sum_{n=0}^\infty \int_0^1 y^{2n} \ln^4 y \ dy\\
=&\frac 3 {16}\sum_{n=1}^\infty \int_0^1 y^{n-1}\ln^4 y \ dy - \frac 3 4 \sum_{n=0}^\infty \frac {24}{(2n+1)^5}\\
=&\frac 3 {16}\sum_{n=1}^\infty \frac{24}{n^5} - 18 \sum_{n=0}^\infty \frac {1}{(2n+1)^5}\\
=&\frac {9}{2} \zeta(5)- 18\cdot \frac {31}{32}\zeta(5)\\
=&\boxed{-\frac{207}{16}\zeta(5)}
\end{align*} as can be found in @Przemo's answer.
For $I_4$, make substitution $ \frac{1-x}{1+x}\mapsto x$ to get
\begin{align*} I_4 = &2\int_0^1 \frac{\ln^3 x \ln\left(\frac 2 {1+x}\right)}{1-x^2} dx \\
=&2\ln 2 \int_0^1 \frac{\ln^3 x}{1-x^2} dx - \underbrace{2\int_0^1\frac{\ln^3 x \ln(1+x)}{1-x^2} dx }_{=:J}\\
=& 2\ln 2\sum_{n=0}^\infty \int_0^1 x^{2n} \ln^3 x\ dx - J\\
=& - 12\ln 2 \underbrace{\sum_{n=0}^\infty \frac 1 {(2n+1)^4}}_{\frac{15}{16}\zeta(4) = \frac{\pi^4}{96}} - J \\
=& -\frac{\pi^4 \ln 2}{8} - J.
\end{align*}
\begin{align*}
J = &\int_0^1\frac{2\ln^3 x \ln(1+x)}{1-x^2} dx \\
=& \underbrace{\int_0^1 \frac{\ln^3 x \ln(1+x)}{1+x}dx}_{=:A} + \int_0^1 \frac{\ln^3 x \ln(1+x)}{1-x}dx\\
=& A + \int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \int_0^1 \frac{(1+x)\ln^3 x \ln(1-x^2)}{1-x^2}dx -\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + \underbrace{\int_0^1 \frac{\ln^3 x \ln(1-x^2)}{1-x^2}dx }_{=:B}+\underbrace{\int_0^1 \frac{x\ln^3 x \ln(1-x^2)}{1-x^2}dx}_{x^2 \mapsto x}-\int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx\\
=&A + B - \underbrace{\frac {15}{16} \int_0^1 \frac{\ln^3 x \ln(1-x)}{1-x}dx}_{=:C}\\
=&A + B - C.
\end{align*}
For $A$, we can use the McLaurin series of
$$
\frac{\ln (1+x)}{1+x} = \sum_{n=0}^\infty (-1)^{n-1}H_n x^n
$$ ($H_0= 0$) to get
\begin{align*}
A = & \sum_{n=0}^\infty (-1)^{n-1}H_n \int_0^1 x^n\ln^3 x \ dx \\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_n}{(n+1)^4}\\
=&6 \sum_{n=0}^\infty \frac{(-1)^{n}H_{n+1}}{(n+1)^4} - 6\sum_{n=0}^\infty \frac{(-1)^{n}}{(n+1)^5}\\
=&6 \sum_{n=1}^\infty \frac{(-1)^{n-1}H_{n}}{n^4} - 6\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^5}\\
=& 6\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-6\cdot \frac{15}{16}\zeta(5)\\
=& \frac{87}{16}\zeta(5) - \frac{\pi^2 \zeta(3)}{2}.
\end{align*}
Here, the known value of $ \sum_{n=1}^\infty (-1)^{n-1}{H_n \over n^4}$ is used.
For $B$, make substitution $u = x^2$ to get
\begin{align*}
B =& \frac 1 {16} \int_0^1 \frac{\ln^3 u \ln(1-u)}{\sqrt u (1-u)} du \\
=& \frac 1 {16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=\frac 1 2, y = 0^+}
\end{align*} where $\text{B}(\cdot,\cdot)$ is Euler's Beta function. We can use the fact that
\begin{align*}
\lim_{y\to 0^+}\frac{\partial^2}{\partial x\partial y} \text{B}(x,y) = -\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]
\end{align*} to get
\begin{align*}
B =& \frac 1 {16}\frac{d^2}{dx^2}\left[-\frac 1 2 \psi''(x) + \psi'(x) \big[\psi(x) + \gamma\big]\right]_{x=\frac 1 2}\\
=&\frac 1 {16} \left[-\frac 1 2 \psi''''(1/2) + \psi'''(1/2)\big[\psi(1/2) + \gamma\big] + 3\psi'(1/2)\psi''(1/2)\right]\\
=& \frac 1 {16}\left[-21\pi^2 \zeta(3) + 372\zeta(5) - 2\pi^4 \ln 2\right]
\end{align*} which can be evaluated using the series representations of polygamma functions $$\psi(x) +\gamma = - \frac 1 x +\sum_{n=1}^\infty \frac 1 n - \frac 1 { n+x},\\
\psi'(x) = \sum_{n=0}^\infty \frac 1 {(n+x)^2}$$ and the derived fact that $\psi(\tfrac 1 2 )+\gamma = -2\ln 2$ and $\psi^{(k)}(\tfrac 1 2)=(-1)^{k+1}k!(2^{k+1}-1)\zeta(k+1)$ for $k\ge 1$.
For $C$, we can use the same method as used in the evaluation of $B$. It holds that
\begin{align*}
C =& \frac {15}{16} \left[\frac{\partial^4}{\partial x^3\partial y} \text{B}(x,y)\right]_{x=1, y = 0^+}\\
=&\frac {15} {16}\left[-\frac 1 2 \psi''''(1) + \psi'''(1)\big[\psi(1) + \gamma\big] + 3\psi'(1)\psi''(1)\right]\\
=&\frac{15}{16}\left[12\zeta(5) -6\zeta(2)\zeta(3)\right]\\
=&\frac {45}{4}\zeta(5) -\frac {15\pi^2 \zeta(3)}{16}
\end{align*} where $\psi(1) +\gamma = 0$, $\psi'(1) = \zeta(2)$, $\psi''(1) = -2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Combining $A,B,C$, we have that $$J =A+B-C= \frac{279}{16}\zeta(5) -\frac{7\pi^2\zeta(3)}{8} - \frac{\pi^4 \ln 2}{8}$$ and
$$
\boxed{I_4 = -\frac{\pi^4 \ln 2}{8} - J = -\frac{279}{16}\zeta(5)+\frac{7\pi^2\zeta(3)}{8}}
$$
Finally, these evaluate $\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =\frac 1 2\big[-I_1-I_2+I_3-I_4\big]$ as follows.
\begin{align*}
\int_0^1 {\ln^3(1-x)\ln(1+x)\over x}dx =&\ 6\text{Li}_5(1/2) + 6\ln 2\ \text{Li}_4(1/2)-\frac{81}{16}\zeta(5)-{7\pi^2 \over 16}\zeta(3)\\
&+\frac{21\ln^2 2}{8}\zeta(3)- \frac{1}{6}\pi^2\ln^3 2+\frac{1}{5}\ln^5 2.
\end{align*}
Using the identity given in the OP, we get the desired integral $I$
\begin{align*}
\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx = &-2\text{Li}_5(1/2) -2\ln 2\ \text{Li}_4(1/2)+\frac{27}{32}\zeta(5) +\frac{7\pi^2}{48}\zeta(3)-\frac{7\ln^2 2}{8}\zeta(3) \\
&-\frac{\pi^4\ln 2}{144} +\frac{\pi^2\ln^3 2}{12} - \frac{7\ln^5 2}{60}.
\end{align*}
I'll just show an idea that avoids those type of sum, but skip the calculations. You might also have better ideas to solve them.
For start we will denote $a=\ln(1-x)$ and $b=\ln(1+x)$ and use the following identity:
$$a^2=\frac12 (a+b)^2+\frac12(a-b)^2-b^2$$
$$\Rightarrow I=\frac12 \underbrace{\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1+x}dx}_{I_1}+\frac12\underbrace{ \int_0^1 \frac{\ln x\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx}_{I_2}-\underbrace{\int_0^1 \frac{\ln x\ln^2(1+x)}{1+x}dx}_{I_3}$$
For the first integral we will write the denominator as:
$$\frac{1}{1+x}=\frac{1}{1-x^2}-\frac{x}{1-x^2}$$
$$\Rightarrow I_1=\int_0^1 \frac{\ln x\ln^2(1-x^2)}{1-x^2}dx-{\int_0^1 \frac{x\ln x\ln^2(1-x^2)}{1-x^2}dx}$$
$$\overset{x^2\to x}=\frac14 \int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}\frac{dx}{\sqrt x}-\frac14\int_0^1 \frac{\ln x\ln^2(1-x)}{1-x}dx$$
Those two integral can be found using the second identity from here.
Let's also take $I_2$ and substitute $\frac{1-x}{1+x}\to x$.
$$\Rightarrow I_2=\underbrace{\int_0^1 \frac{\ln(1-x)\ln^2 x}{1+x}dx}_{P}-\underbrace{\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx}_{Q}$$
$$P-Q=I_2;\quad P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1+x}dx$$
And again with the same trick done for $I_1$, we have:
$$P+Q=\int_0^1 \frac{\ln(1-x^2)\ln^2 x}{1-x^2}dx-\int_0^1 \frac{x\ln(1-x^2)\ln^2 x}{1-x^2}dx$$
$$\overset{x^2\to x}=\frac18\int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}\frac{dx}{\sqrt x}-\frac18 \int_0^1 \frac{\ln(1-x)\ln^2 x}{1-x}dx$$
Henceforth we can extract our second integral, $I_2$ as:
$$I_2=P-Q=(P+Q)-2Q$$
Note that $P+Q$ can again be found using the second identity from here.
Finally, we only need to find $Q$.
$$Q=\int_0^1 \frac{\ln(1+x)\ln^2 x}{1+x}dx=\sum_{n=1}^\infty (-1)^{n+1} H_n\int_0^1 x^{n}\ln^2 x=2\sum_{n=1}^\infty \frac{(-1)^{n+1}H_n}{(n+1)^3}$$
So $Q$ is actually an Euler sum in disguise, but you nicely found it here.
Also, $I_3$ is pretty easy, one just needs to use the same approach as done for $I_1$ in your following post.
$$I_3=\int_0^1 \frac{\ln x \ln^2(1+x)}{1+x}dx\overset{IBP}=-\frac12\int_0^1 \frac{\ln^3(1+x)}{x}dx$$
Best Answer
Starting with the following identity that can be found on page $95$ Eq $(5)$ of this paper
$$\sum_{n=1}^\infty \overline{H}_n\frac{x^n}{n}=\operatorname{Li}_2\left(\frac{1-x}{2}\right)-\operatorname{Li}_2(-x)-\ln2\ln(1-x)-\operatorname{Li}_2\left(\frac12\right)$$
Multiply both sides by $\large \frac{\ln(1-x^2)}{x}$ then integrate from $x=0$ to $x=1$ we get
$$\underbrace{\sum_{n=1}^\infty \frac{\overline{H}_n}{n}\int_0^1x^{n-1}\ln(1-x^2)\ dx}_{\large S}$$ $$\small{=I-\underbrace{\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x^2)}{x}\ dx}_{\large J}-\ln2\underbrace{\int_0^1\frac{\ln(1-x)\ln(1-x^2)}{x}\ dx}_{\large K}-\operatorname{Li}_2\left(\frac12\right)\underbrace{\int_0^1\frac{\ln(1-x^2)}{x}\ dx}_{\large -\frac12\zeta(2)}}$$
or
$$I=S+J+\ln2\ K-\frac12\zeta(2)\operatorname{Li}_2\left(\frac12\right)\tag1$$
Evaluating $S$
Notice that
$$\int_0^1 x^{n-1}\ln(1-x^2)\ dx\overset{x^2\to x}{=}\frac12\int_0^1 x^{n/2-1}\ln(1-x)\ dx=-\frac{H_{n/2}}{n}$$
$$\Longrightarrow S=\boxed{-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}}$$
Evaluating $J$
Writing $\ln(1-x^2)=\ln(1-x)+\ln(1+x)$ gives
$$J=\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1-x)}{x}dx+\int_0^1\frac{\operatorname{Li}_2(-x)\ln(1+x)}{x}dx$$
$$=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\int_0^1 x^{n-1}\ln(1-x)\ dx-\frac12\operatorname{Li}_2^2(-x)|_0^1$$
$$=-\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}-\frac{5}{16}\zeta(4)$$
$$=\boxed{-2\operatorname{Li_4}\left(\frac12\right)+\frac{39}{16}\zeta(4)-\frac74\ln2\zeta(3)+\frac12\ln^22\zeta(2)-\frac{1}{12}\ln^42}$$
where we used $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^3}$$=2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42$
Evaluating $K$
Similarly
$$K=\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}\ dx}_{\large 2\zeta(3)}+\underbrace{\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}\ dx}_{\large -\frac{5}{8}\zeta(3)}=\boxed{\frac{11}8\zeta(3)}$$
where the second integral is evaluated here.
Plug the boxed results in $(1)$ we obtain that
$$I=-2\operatorname{Li}_4\left(\frac12\right)+\frac{29}{16}\zeta(4)-\frac38\ln2\zeta(3)+\frac34\ln^22\zeta(2)-\frac1{12}\ln^42-\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}$$
In here we proved
$$\sum_{n=1}^\infty\frac{\overline{H}_nH_{n/2}}{n^2}=\frac1{24}\ln^42-\frac14\ln^22\zeta(2)+\frac{21}{8}\ln2\zeta(3)-\frac{9}{8}\zeta(4)+\operatorname{Li}_4\left(\frac12\right)$$