Using the generating function of $\displaystyle\{H_k^2\}_{k=1}^\infty$:
$$
\frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} = \sum_{k=1}^\infty H_k^2 x^k
$$ we can observe that
\begin{align*}
S =& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1{k^2}\\
=& \sum_{k=1}^\infty {H_k^2 \over 2^k} \frac 1 2\int_0^1 x^{k-1}\ln^2 x\ dx\\
=&\frac 1 2 \int_0^1 \left(\sum_{k=1}^\infty H_k^2\left(\frac x 2\right)^k\right)\frac{\ln^2 x}{x} dx\\
=& \frac 1 2 \int_0^{\frac 1 2}\left(\sum_{k=1}^\infty H_k^2 x ^k\right) \frac{\ln^2 (2x)}{x} dx \\
=& \frac 1 2 \int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln^2 x}{x} dx \\
&+ \ln 2\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{\ln x}{x} dx \\
&+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x) +\ln^2(1-x)}{1-x} \frac{dx}{x}\\
=&: I_1 + I_2 + I_3.
\end{align*}
For $I_1$, we have
\begin{align*}
I_1=&\frac 1 2 \int_0^{\frac 12 } \frac{\big[\text{Li}_2(x)+\ln^2(1-x)\big]\ln^2 x}{x(1-x)}dx \\
=& \frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{x}dx+\frac 1 2 \int_0^{\frac 12 } \frac{\text{Li}_2(x)\ln^2 x}{1-x}dx +\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2 (1-x)\ln^2 x}{x(1-x)} dx\\
=&:I_1'+I_1''+I_1'''.
\end{align*}
For $I_1'$, we integrate by parts twice to get
\begin{align*}
I_1' \underset{\text{IBP}}{=}& \frac 1 2\left[ \text{Li}_3(x)\ln^2 x\right]^{1/2}_0 -\int_0^{\frac 1 2}\frac{\text{Li}_3(x) \ln x}{x} dx\\
\underset{\text{IBP}}{=}&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 - \left[\text{Li}_4(x)\ln x\right]^{1/2}_0 +\int_0^{\frac 1 2 }\frac{\text{Li}_4(x)}x dx\\
=&\frac{\ln^2 2\ \text{Li}_3(1/2)}2 + \ln 2\ \text{Li}_4(1/2)+\text{Li}_5(1/2)\\
=&\boxed{\text{Li}_5(1/2)+\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{16}\zeta(3)-\frac {\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{12}}
\end{align*} where the well-known value of
$
\text{Li}_3(1/2) = \frac 78 \zeta(3) -\frac{\pi^2\ln 2}{12}+\frac{\ln^3 2}{6}
$ is used to simplify.
For $I_1''$, by integrating by parts,
\begin{align*}
I_1'' \underset{\text{IBP}}{=}& \frac 1 2 \int_0^{\frac 1 2} \ln(1-x)\left[\frac{2\ln x\text{Li}_2(x)}{x} - \frac{\ln(1-x)\ln^2 x}{x}\right]dx +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
=& {\int_0^{\frac 1 2} \ln x\frac{\ln(1-x)\text{Li}_2(x)}{x} dx}-\underbrace{\frac 1 2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln^2 x}{x}dx}_{=:J} +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
\underset{\text{IBP}}{=}&\frac{\ln 2}2\text{Li}_2^2(1/2) +{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J +\frac{\ln^3 2}{2}\text{Li}_2(1/2)\\
=&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+{\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx}-J.
\end{align*}
The well-known value of $\text{Li}_2(1/2) = \frac{\pi^2}{12} - \frac{\ln^2 2}{2}$ is used to simplify. In fact, the integral ${\int_0^{1/2}\frac{\text{Li}_2^2(x)}{x} dx}$ was already evaluated in my previous answer here:
\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*}
For $J$, we make substitution $y= \frac{x}{1-x}$ to get
\begin{align*}
J=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y(1+y)}dy\\
=&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy-\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\
=&:K-L.
\end{align*}
For $K$, expanding $\ln^2\left(\frac y {1+y}\right)=\big[\ln y -\ln(1+y)\big]^2$ and integrating by parts we get
\begin{align*}
K =&\frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{y}dy\\
=& \frac 12{ \int_0^1\frac{\ln^2 y\ln^2(1+y)}{y} dy}-{\int_0^1 \frac{\ln y\ln^3(1+y)}{y} dy}+\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy\\
\underset{\text{IBP}}{=}&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \frac 3 2\int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{y}dy
\end{align*} Doing the same thing for $L$,
\begin{align*}
L = & \frac 1 2\int_0^1 \frac{\ln^2(1+y)\ln^2\left(\frac{y}{1+y}\right)}{1+y}dy\\
=& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy- {\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy} +\frac 12 \int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\
\underset{\text{IBP}}{=}& \frac 12 \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy+\frac 1 4 \int_0^1 \frac{\ln^4(1+y)}y dy +\frac{\ln^5 2}{10}.
\end{align*} This gives that
\begin{align*}
J=&K-L\\
=&-\frac 1 3 \int_0^1 \frac{\ln^3 y \ln(1+y)}{1+y}dy + \int_0^1 \frac{\ln^2 y\ln^2(1+y)}{1+y}dy +\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy -\frac{\ln^5 2}{10}\\
=&:-V_1+V_2+V_3 -\frac{\ln^5 2}{10}.
\end{align*}
For $V_1$, we can use the Maclaurin series of $\frac{\ln (1+y)}{1+y} = \sum_{k=0}^\infty (-1)^{k-1} H_k y^k$ to get
\begin{align*}
V_1=&\frac 1 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy \\
=& \frac 1 3\sum_{k=0}^\infty (-1)^{k-1}H_k {\int_0^1 y^k\ln^3 y\ dy} \\
=& \frac{-6}{3}\sum_{k=0}^\infty \frac{(-1)^{k-1}H_k}{(k+1)^4}\\
=&2\sum_{k=0}^\infty \frac{(-1)^k \left(H_{k+1}-\frac 1{k+1}\right)}{(k+1)^4} \\
=&2 \sum_{k=1}^\infty \frac{(-1)^{k-1}H_k}{k^4} -2 \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^5}\tag{$k+1\mapsto k$}\\
=&2\left(\frac{59}{32}\zeta(5) - \frac{\pi^2\zeta(3)}{12}\right)-2\cdot \frac{15}{16}\zeta(5)\\
=&\frac{29}{16}\zeta(5) - \frac{\pi^2}{6}\zeta(3)
\end{align*} where the known value of alternating Euler sum $\sum_{k=1}^\infty \tfrac{(-1)^{k-1}H_k}{k^4}$ is used.
For $V_2$, we consider the algebraic identity
$$
6a^2b^2 = (a-b)^4 - a^4 +4a^3b +4ab^3 -b^4
$$ with $a=\ln y$ and $b = \ln(1+y)$ to get
\begin{align*}
V_2 =&\frac 1 6{\int_0^1 \frac{\ln^4\left(\frac y{1+y}\right)}{1+y} dy}-\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +\underbrace{\frac 2 3\int_0^1 \frac{\ln^3 y\ln(1+y)}{1+y} dy}_{=2V_1}\\
&+\frac 2 3\underbrace{\int_0^1 \frac{\ln y\ln^3(1+y)}{1+y} dy}_{=- V_3\text{ by IBP}} -\frac 1 6\int_0^1 \frac{\ln^4(1+y)}{1+y} dy\\
=&\frac 1 6 \int_0^{\frac 1 2} \frac{\ln^4 x}{1-x} dx -\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy +2V_1-\frac 2 3 V_3 -\frac{\ln^5 2}{30}.\tag{$\tfrac y{1+y}= x$}
\end{align*}
For the first integral, we have
\begin{align*}
W:=&\frac 1 6{\int_0^{\frac 1 2 }\frac{\ln^4 x}{1-x} dx}\tag{$2x\mapsto x$} \\
=& \frac 1 6\int_0^1 \frac{\ln^4(\tfrac x 2)}{2-x}dx\\
=&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\int_0^1 x^{k-1}\Big[\ln^4 x -4\ln 2 \ln^3 x + 6\ln^2 2\ln^2 x - 4\ln^3 2 \ln x + \ln^4 2\Big]dx\\
=&\frac 1 6\sum_{k=1}^\infty \frac{1}{2^k}\left[\frac{24}{k^5} + \frac{24\ln 2}{k^4} +\frac {12\ln^2 2}{k^3} +\frac{4\ln^3 2}{k^2} +\frac{\ln^4 2}{k}\right]\\
=&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + 2\ln^2 2\text{Li}_3(1/2) + \frac{2\ln^3 2}{3}\text{Li}_2(1/2) + \frac{\ln^5 2}6\\
=&4\text{Li}_5(1/2) +4\ln 2\text{Li}_4(1/2) + \frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^2\ln^3 2}{9}+ \frac{\ln^5 2}6.
\end{align*}
For the second integral, we have
\begin{align*}
\frac 1 6\int_0^1 \frac{\ln^4 y}{1+y} dy
=& \frac 1 6\sum_{k=1}^\infty (-1)^{k-1} \int_0^1 y^{k-1}\ln^4 y \ dy \\
=& \frac 1 6 \sum_{k=1}^\infty(-1)^{k-1} \frac{24}{k^5}\\
=&\frac{15}{4}\zeta(5).
\end{align*}
This gives
$$
V_2 = W +2V_1-\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}.
$$
For $V_3$ we have
\begin{align*}
V_3=&\frac 14 \int_0^1 \frac{\ln^4(1+y)}{y}dy \tag{$y\mapsto y+1$}\\
=& \frac 1 {4}{ \int_1^2 \frac{\ln^4 y }{y-1} dy} \tag{$\tfrac 1 y\mapsto y$}\\
=&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y(1-y)} dy\\
=&\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{y} dy+\frac 1 {4} \int_{\frac 1 2}^1 \frac{\ln^4 y}{1-y} dy\\
=&\frac{\ln^5 2}{20} + \frac 1 {4} \int_0^1 \frac{\ln^4 y}{1-y} dy -\frac 1 {4} \underbrace{\int_0^{\frac 1 2}\frac{\ln^4 y}{1-y}dy}_{=6 W}\\
=&\frac{\ln^5 2}{20} +\frac 1 {4} \sum_{k=1}^\infty \int_0^1 y^{k-1}\ln^4 y\ dy- \frac 3 2 W\\
=&\frac{\ln^5 2}{20} +6\zeta(5)- \frac 3 2 W.
\end{align*}
Combining $V_1$, $V_2$ and $V_3$, we get
\begin{align*}
J = & V_2 -V_1+V_3 -\frac{\ln^5 2}{10}\\
=& \left[W+2V_1 -\frac 2 3 V_3-\frac{15}{4}\zeta(5)-\frac{\ln^5 2}{30}\right]-V_1+V_3 -\frac{\ln^5 2}{10}\\
=& W+V_1+\frac 1 3 V_3-\frac{15}{4}\zeta(5)-\frac{2\ln^5 2}{15}\\
=&\frac 1 2 W+V_1 -\frac 7 4\zeta(5) -\frac{7\ln^5 2}{60}\\
=&2\text{Li}_5(1/2) +2\ln 2\ \text{Li}_4(1/2) +\frac 1 {16}\zeta(5) -\frac{\pi^2}6 \zeta(3) +\frac {7\ln^2 2}{8} \zeta(3) -\frac{\ln^2 2\pi^3}{18}-\frac{\ln^5 2}{30}.
\end{align*}
This gives
\begin{align*}
I_1'' =&\frac{\pi^4\ln 2}{288} -\frac{\ln^5 2}{8}+\frac 1 2\int_0^{\frac 1 2}\frac{\text{Li}_2^2(x)}{x} dx-J\\
=&\boxed{\small -3\text{Li}_5(1/2) -3\ln 2\text{Li}_4(1/2) +\frac{23}{64}\zeta(5) +\frac {23\pi^2}{96}\zeta(3) -\frac {21\ln^2 2}{16}\zeta(3) +\frac{7\pi^2\ln^3 2}{72} - \frac{3\ln^5 2}{20}.}
\end{align*}
For $I_1'''$, we exploit the symmetric nature of the integrand to write
\begin{align*}
I_1''' :=& \frac 1 2\int_0^{\frac 1 2} \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx\\
=& \frac 1 4\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x(1-x)} dx \\
=& \frac 1 4\underbrace{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{x} dx}_{1-x\mapsto x}+\frac 1 4{\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx}\\
=&\frac 1 2\int_0^1 \frac{\ln^2 x \ln^2 (1-x)}{1-x} dx\\
=&\frac 1 2 \left[\frac{\partial^4}{\partial x^2 \partial y^2 } \text{B}(x,y)\right]_{x=1,y=0^+}
\end{align*} where $\text{B}(x,y)=\tfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ is Euler's Beta function. Now we can use the fact that
\begin{align*}
\lim_{y\to 0^+}\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,y)
=&-\frac 1 3\psi'''(x)+\psi''(x)\Big[\psi(x) +\gamma\Big] + \psi'(x)\Big[\psi'(x)-\zeta(2) - \big[\psi(x) + \gamma\big]^2\Big]
\end{align*} to obtain
\begin{align*}
I_1'''=& \frac 1 2\frac d{dx}\left[\frac{\partial^3 }{\partial x\partial y^2 }\text{B}(x,0^+)\right]_{x=1} \\
=& -\frac 1 6 \psi''''(1) +\psi'(1)\psi''(1) \\
=&\boxed{4\zeta(5) -\frac{\pi^2}3 \zeta(3)}
\end{align*} where the values of $\psi(1) +\gamma = 0$, $\psi'(1) =\zeta(2)$, $\psi''(1) =-2\zeta(3)$ and $\psi''''(1) = -24\zeta(5)$ are used.
Finally, from $I_1 = I_1'+I_1''+I_1'''$ we get
\begin{align*}
I_1 =& -2\text{Li}_5(1/2) - 2\ln 2\text{Li}_4(1/2) + \frac {279}{64}\zeta(5) -\frac {3\pi^2}{32}\zeta(3)-\frac {7\ln^2 2}{8} \zeta(3)+\frac {\pi^2\ln^3 2}{18}-\frac{\ln^5 2}{15}.
\end{align*}
For $I_2$, we observe that
\begin{align*}\require{cancel}
I_2 =& \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x(1-x)} dx \\
=& \ln 2 {\int_0^{\frac 12} \frac{\text{Li}_2(x) \ln x}{1-x} dx}+\ln 2\int_0^{\frac 12} \frac{\ln^2(1-x)\ln x}{1-x} dx+ \ln 2 \int_0^{\frac 12} \frac{\big[\text{Li}_2(x) + \ln^2(1-x)\big]\ln x}{x} dx\\
\underset{\text{IBP}}{=}&-\ln^3 2\ \text{Li}_2(1/2)+\ln 2 \int_0^{\frac 12} \ln(1-x)\frac{-\cancel{\ln(1-x)\ln x}+\text{Li}_2(x)}{x} dx \\
&+\ln2 \int_0^{\frac 1 2} \frac{\ln^2(1-x)\ln x}{1-x} dx + \ln 2{ \int_0^{\frac 12} \frac{\big[\text{Li}_2(x)+\cancel{ \ln^2(1-x)}\big] \ln x}{x} dx}\\
\underset{\text{IBP}}{=}&\small-\ln^3 2 \text{Li}_2(1/2)-\tfrac{\ln 2}{2} \left[\text{Li}^2_2(x)\right]^{1/2}_0-\frac{\ln^5 2}3+{\frac{\ln 2}3{\int_0^{\frac 1 2} \frac{\ln^3(1-x)}{x} dx}} -\ln^2 2\ \text{Li}_3(1/2)-\ln 2\int_0^{\frac 1 2} \frac{\text{Li}_3(x)}{x} dx\normalsize\\
=&-\ln^3 2\ \text{Li}_2(1/2) -\tfrac{\ln 2}{2} \text{Li}^2_2(1/2)-\frac{\ln^5 2}3+ \small\underbrace{\frac{\ln 2}3{\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx}}_{1-x\mapsto x, \ =:I_2'}\normalsize-\ln^2 2\ \text{Li}_3(1/2)-\ln 2\ \text{Li}_4(1/2)\\
=&-\ln 2\ \text{Li}_4(1/2) -\frac{7\ln^2 2}8 \zeta(3) -\frac{\pi^4 \ln 2}{288}+\frac {\pi^2\ln^3 2}{24} -\frac{\ln^5 2}{8} + I_2'.
\end{align*}
For $I_2'$, by integrating by parts, we have
\begin{align*}
I_2'
=& \frac{\ln 2}3\int_{\frac 1 2}^1 \frac{\ln^3 x}{1-x} dx\\
=&\frac{\ln 2}3\int_{0}^1 \frac{\ln^3 x}{1-x} dx -\frac{\ln 2}3{\int_{0}^{\frac 12} \frac{\ln^3 x}{1-x} dx}\tag{$x=\tfrac y 2$}\\
=&\frac{\ln 2}3\sum_{k=1}^\infty {\int_0^1 x^{k-1}\ln^3 x\ dx}-\underbrace{\frac{\ln 2}3\int_{0}^{1} \frac{\ln^3 (\tfrac y 2)}{2-y} dy}_{=:A}\\
=&-\frac{\pi^4\ln 2}{45}-A.
\end{align*}
\begin{align*}
A=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\ln^3 (\tfrac y 2) dy\\
=&\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\int_{0}^{1} y^{k-1}\left[\ln^3 y - 3\ln 2\ln^2 y +3\ln^2 2\ln y -\ln^3 2\right]dy\\
=&-\frac{\ln 2}3\sum_{k=1}^\infty \frac 1 {2^k}\left[\frac 6 {k^4}+\frac{6\ln 2}{k^3} +\frac{3\ln^2 2}{k^2} +\frac{\ln^3 2}{k}\right]\\
=& -2\ln 2\ \text{Li}_4(1/2) - 2\ln^2 2\ \text{Li}_3(1/2)-\ln^3 2\ \text{Li}_2(1/2)-\frac{\ln^5 2}3\\
=&-2\ln 2\ \text{Li}_4(1/2)-\frac{7\ln^2 2}{4}\zeta(3)+\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}.
\end{align*}
This gives
$$
I_2'= 2\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{4}\zeta(3)-\frac{\pi^4\ln 2}{45}-\frac{\pi^2\ln^3 2}{12}-\frac{\ln^5 2}{6}
$$ and
\begin{align*}
I_2=\ln 2\ \text{Li}_4(1/2)+\frac{7\ln^2 2}{8}\zeta(3)-\frac{37\pi^4\ln 2}{1440}-\frac{\pi^2\ln^3 2}{24}+\frac{\ln^5 2}{24}.
\end{align*}
For $I_3$, we have
\begin{align*}
I_3=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)+\ln^2(1-x)}{x(1-x)}dx\\
=&\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\text{Li}_2(x)}{x}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} {\left[\frac{\ln^2(1-x)}{x}+\frac{\text{Li}_2(x)}{1-x}\right]}dx+\frac{\ln^2 2}{2}\int_0^{\frac 1 2} \frac{\ln^2(1-x)}{1-x}dx\\
=&\frac{\ln^2 2}{2}\text{Li}_3(1/2) +\frac{\ln^2 2}{2}\big[-\ln(1-x)\text{Li}_2(x)\big]^{1/2}_0+\frac{\ln^5 2}6.
\end{align*} Using the well-known values of $\text{Li}_3(1/2)$ and $\text{Li}_2(1/2)$, this simplifies to
$$
I_3 =\frac {7\ln^2 2}{16}\zeta(3).
$$
From $S = I_1+I_2 + I_3$, we finally get
\begin{align*}
\sum_{k=1}^\infty \frac{H_k^2}{k^32^k} =& -2\text{Li}_5(1/2) -\ln 2\ \text{Li}_4(1/2) + \frac{279}{64}\zeta(5) - \frac{3\pi^2}{32}\zeta(3) + \frac{7\ln^2 2}{16}\zeta(3) - \frac{37\pi^4 \ln 2}{1440}\\
& + \frac{\pi^2 \ln^3 2}{72} - \frac{\ln^5 2}{40}.
\end{align*}
We can observe that the values of $I_2$ and $I_3$ can be used to evaluate sums of lower order in a similar way:
\begin{eqnarray*}
&\sum_{k=1}^\infty \frac{H_k^2}{k^2 2^k} = -\frac 1{\ln 2} I_2 - \frac{2}{\ln 2} I_3=-\text{Li}_4(1/2) -\frac{7\ln 2}4\zeta(3) +\frac{37 \pi^4}{1440} +\frac{\pi^2\ln^2 2}{24} -\frac{\ln^4 2}{24},\\
&\sum_{k=1}^\infty \frac{H_k^2}{k 2^k} = \frac{2}{\ln^2 2} I_2 = \frac{7}{8}\zeta(3).
\end{eqnarray*}
A second solution in large steps by Cornel Ioan Valean
Let's start with the following useful identity which is easily derived by using recurrence relations and simple rearrangements, manipulations with sums, that is
Let $n$ be a non-negative integer number. Then, we have
$$\int_0^1 x^{2n}\frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=\frac{1}{2}H_{2n}^2-2\log(2) H_{2n}+\frac{1}{2}H_{2n}^{(2)}-\frac{1}{4}H_n^2-\frac{1}{4}H_n^{(2)}+\log (2)H_n+\frac{1}{2} \log ^2(2)-\sum_{k=1}^{n-1}\frac{H_k}{2 k+1},$$
where $H_n^{(m)}=1+\frac{1}{2^m}+\cdots+\frac{1}{n^m}$ represents the $n$th generalized harmonic number of order $m$.
By multiplying both sides of the identity above by $1/n^3$ and considering the summation from $n=1$ to $\infty$, we get
$$\sum_{n=1}^{\infty} \frac{1}{n^3}\sum_{k=1}^{n-1}\frac{H_{k}}{2 k+1}=\sum_{k=1}^{\infty} \sum_{n=k+1}^{\infty}\frac{1}{n^3}\frac{H_{k}}{2 k+1}=\underbrace{\sum_{k=1}^{\infty}\frac{H_{k}}{2 k+1}\left(\zeta(3)-H_k^{(3)}\right)}_{\text{The desired series}}$$
$$=\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^2}{n^3}-2\log(2) \sum_{n=1}^{\infty}\frac{H_{2n}}{n^3}+\frac{1}{2}\sum_{n=1}^{\infty}\frac{H_{2n}^{(2)}}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{H_n^2}{n^3}-\frac{1}{4}\sum_{n=1}^{\infty} \frac{H_n^{(2)}}{n^3}$$
$$+\log (2)\sum_{n=1}^{\infty} \frac{H_n}{n^3}+\frac{1}{2}\log ^2(2)\sum_{n=1}^{\infty}\frac{1}{n^3}-\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x,$$
where we see all the series in the right-hand side are easily reducible to known series which may also be found in the book (Almost) Impossible Integrals, Sums, and Series.
On the other hand, with simple integration by parts, we obtain
$$\int_0^1 \frac{\log(1+x)}{1+x}\operatorname{Li}_3(x^2)\textrm{d}x$$
$$=\frac{1}{2}\log^2(2)\zeta(3)-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(x)}{x}\textrm{d}x-2\int_0^1 \frac{\log^2(1+x)\operatorname{Li}_2(-x)}{x}\textrm{d}x,$$
where the last integrals may be found calculated in the paper The calculation of a harmonic series with a weight $5$ structure, involving the product of harmonic numbers, $H_n H_{2n}^{(2)}$.
A note: The sister of the result above (easy to obtain by recurrence relations and very useful),
$$\int_0^1 x^{2n-1} \frac{\log(1+x)}{1+x}\textrm{d}x$$
$$=2\log(2) H_{2n}-\log(2)H_n+\frac{1}{4}H_n^2+\frac{1}{4}H_n^{(2)}-\frac{1}{2}H_{2n}^2-\frac{1}{2} H_{2n}^{(2)}+\frac{H_{2n}}{2n}-\frac{H_n}{2n}
$$
$$
-\frac{1}{2}\log^2(2)+\sum_{k=1}^{n-1}\frac{H_k}{2 k+1}.
$$
Best Answer
First, we prove a lemma on the integral representation of $(H_n)^2$. $$I_n=\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)=(H_n)^2$$
Let's prove by induction. $\displaystyle I_0=-\int_0^1\frac{\ln x}{1-x}dx=\zeta(2)=\zeta(2)+(H_0)^2$.\ Assume the equation holds for $n-1$, $$\begin{aligned} I_n&=\int_0^1\left(2(x^n-1)\frac{\ln(1-x)}{1-x}-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\\ &=I_{n-1}+\int_0^1\left(2(x^n-x^{n-1})\frac{\ln(1-x)}{1-x}-\frac{(x^n-x^{n-1})\ln x}{1-x}\right)d x\\ &=(H_{n-1})^2+\int_0^1\left(-2x^{n-1}\ln(1-x)+x^{n-1}\ln x\right)d x\\ &=\left(H_n-\frac1n\right)^2-\frac1{n^2}+2\cdot\frac{H_n}n=(H_n)^2 \end{aligned}$$ Result Therefore, and by integrating $\displaystyle\sum_{n=1}^\infty\frac{\binom{2n}n}{4^n}x^n=\frac{1}{\sqrt{1-x}}-1$ from $0$ with respect to $x$, we have $$\begin{aligned} S&=\sum_{n=1}^\infty\frac1n\frac{\binom{2n}n}{4^n}\left(\int_0^1\left(nx^{n-1}\ln^2(1-x)-\frac{x^n\ln x}{1-x}\right)d x-\zeta(2)\right)\\ &=\int_0^1\left(\frac{1}{x\sqrt{1-x}}-\frac1x\right)\ln^2(1-x)d x-\int_0^12\ln\frac{2}{1+\sqrt{1-x}}\frac{\ln x}{1-x}d x-2\ln2\zeta(2)\\ &=I_1-I_2-2\ln2\zeta(2) \end{aligned}$$ $I_1=12\zeta(3)$ can be easily deduced by substitution $x\mapsto 1-x^2$. $-2\ln2\zeta(2)+\frac32\zeta(3)$, the value of $I_2$, can also be deduced by the same substitution. By combining these results, $S=\frac{21}2\zeta(3)$.