starting with the integral representation of $$\displaystyle H_n^{(2)}=\zeta(2)+\int_0^1\frac{t^n\ln t}{1-t}\ dt\tag1$$ we can write our sum:
\begin{align}
S&=\sum_{n=1}^\infty\frac{\left(H_n^{(2)}\right)^2}{2^n}=\sum_{n=1}^\infty\frac1{2^n}\left(\zeta(2)+\int_0^1\frac{x^n\ln x}{1-x}\ dx\right)\left(\zeta(2)+\int_0^1\frac{y^n\ln y}{1-y}\ dy\right)\\
&=\small{\sum_{n=1}^\infty\frac1{2^n}\left(\zeta^2(2)+\zeta(2)\int_0^1\frac{y^n\ln y}{1-y}\ dy+\zeta(2)\int_0^1\frac{x^n\ln x}{1-x}\ dx+\int_0^1\int_0^1\frac{(xy)^n\ln x\ln y}{(1-x)(1-y)}\ dx\ dy\right)}
\end{align}
note that the second and the third term have the same value and using the geometric series, we have
\begin{align}
S&=\zeta^2(2)+2\zeta(2)\int_0^1\frac{\ln x}{1-x}\sum_{n=1}^\infty\left(\frac{x}{2}\right)^n\ dx+\int_0^1\int_0^1\frac{\ln x\ln y}{(1-x)(1-y)}\sum_{n=1}^\infty\left(\frac{xy}{2}\right)^n\ dx\ dy\\
&=\zeta^2(2)+2\zeta(2)\int_0^1\frac{x\ln x}{(1-x)(2-x)}\ dx+\int_0^1\int_0^1\frac{xy\ln x\ln y}{(1-x)(1-y)(2-xy)}\ dx\ dy\\
&=\zeta^2(2)+2\zeta(2)(-\ln^22)+\int_0^1\frac{\ln x}{1-x}\left(\int_0^1\frac{xy\ln y}{(1-y)(2-xy)}\ dy\right)\ dx\\
&=\zeta^2(2)-2\zeta(2)\ln^22+\int_0^1\frac{\ln x}{(1-x)(2-x)}\left(\int_0^1\frac{x\ln y}{1-y}\ dy-\int_0^1\frac{2x\ln y}{2-xy}\ dy\right)\ dx\\
&=\zeta^2(2)-2\zeta(2)\ln^22+\int_0^1\frac{\ln x}{(1-x)(2-x)}\left(-\zeta(2)x+2\operatorname{Li_2}\left(\frac{x}{2}\right)\right)\ dx\\
&=\zeta^2(2)-2\zeta(2)\ln^22+(-\ln^22)(-\zeta(2))+2\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx\\
&=\zeta^2(2)-\zeta(2)\ln^22+2\color{blue}{\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx}\\
&=\frac52\zeta(4)-\zeta(2)\ln^22+2\left(\color{blue}{\operatorname{Li_4}\left(\frac{1}{2}\right)-\frac98\zeta(4)+\frac12\ln2\zeta(3)+\frac1{12}\ln^42}\right)\\
&=2\operatorname{Li_4}\left(\frac{1}{2}\right)+\frac14\zeta(4)+\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\ln^42
\end{align}
Evaluation of the blue integral:
$$\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{(1-x)(2-x)}\ dx=\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{1-x}\ dx-\frac12\int_0^1\frac{\ln x\operatorname{Li_2}\left(\frac{x}{2}\right)}{1-\frac x2}\ dx$$
expand $\text{Li}_2(x/2)$ in series in the first integral and use that $\sum_{n=1}^\infty H_n^{(2)}x^n=\frac{\text{Li}_2(x)}{1-x}$ for the second integral with replacing $x$ by $x/2$
$$=\sum_{n=1}^\infty\frac{1}{n^22^n}\int_0^1\frac{x^n \ln x}{1-x}dx-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^n}\int_0^1 x^n\ln xdx$$
use $(1)$ for the first integral
$$=\sum_{n=1}^\infty\frac{1}{n^22^n}\left(-\zeta(2)+H_n^{(2)}\right)-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{2^n}\left(-\frac1{(n+1)^2}\right)$$
reindex the second sum
$$=-\zeta(2)\text{Li}_2(1/2)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}+\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^22^n}$$
write $H_{n-1}^{(2)}=H_n^{(2)}-\frac1{n^2}$ and simplify
$$=-\zeta(2)\text{Li}_2(1/2)+2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}-\sum_{n=1}^\infty\frac{1}{n^42^n}$$
$$=-\text{Li}_4(1/2)-\zeta(2)\text{Li}_2(1/2)+2\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^22^n}$$
substitute
\begin{align*}
\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{{n^22^n}}=\operatorname{Li_4}\left(\frac12\right)+\frac1{16}\zeta(4)+\frac14\ln2\zeta(3)-\frac14\ln^22\zeta(2)+\frac1{24}\ln^42
\end{align*}
and $\text{Li}_2(1/2)=\frac12\zeta(2)-\frac12\ln^2(2)$ , the blue closed form follows.