\begin{align}
S_1&=\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=\sum_{n=1}^\infty\frac1{n^7}\left(\zeta(2)-\sum_{k=1}^\infty\frac1{(n+k)^2}\right)\\
&=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{1}{n^7(n+k)^2}\right)\\
&\small{=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\sum_{n=1}^\infty\frac{7}{k^8}\left(\frac1n-\frac1{n+k}\right)-\frac{6}{k^7n^2}-\frac{1}{k^7(n+k)^2}+\frac{5}{k^6n^3}-\frac{4}{k^5n^4}+\frac{3}{k^4n^5}-\frac{2}{k^3n^6}+\frac{1}{k^2n^7}\right)}\\
&\small{=\zeta(2)\zeta(7)-\sum_{k=1}^\infty\left(\frac{7H_k}{k^8}-\frac{6\zeta(2)}{k^7}-\frac1{k^7}\left(\zeta(2)-H_k^{(2)}\right)+\frac{5\zeta(3)}{k^6}-\frac{4\zeta(4)}{k^5}+\frac{3\zeta(5)}{k^4}-\frac{2\zeta(6)}{k^3}+\frac{\zeta(7)}{k^2}\right)}\\
&=\zeta(2)\zeta(7)-7\sum_{k=1}^\infty\frac{H_k}{k^8}+6\zeta(2)\zeta(7)-S_1-3\zeta(3)\zeta(6)+\zeta(4)\zeta(5)\\
2S_1&=7\zeta(2)\zeta(7)-3\zeta(3)\zeta(6)+\zeta(4)\zeta(5)-7\sum_{k=1}^\infty\frac{H_k}{k^8}
\end{align}
and by substituting $\displaystyle\sum_{k=1}^\infty\frac{H_k}{k^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(3)\zeta(6)-\zeta(4)\zeta(5)$, we get the closed form of $S_1$.
From here, we have
$$\int_0^1x^{n-1}\ln^2(1-x)\ dx=\frac{H_n^2+H_n^{(2)}}{n}$$
Divide both sides by $n^6$ then sum both sides from $n=1$ to $\infty$ to get
\begin{align}
S_2+S_1&=\sum_{n=1}^\infty\frac{H_n^2}{n^7}+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^7}=\int_0^1\frac{\ln^2(1-x)}{x}\sum_{n=1}^\infty\frac{x^n}{n^6}\ dx\\
&=\int_0^1\frac{\ln^2(1-x)\operatorname{Li}_6(x)}{x}\ dx,\quad \left\{\color{red}{\text{use}\ \ln^2(1-x)=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n}\right\}\\
&=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1} \operatorname{Li}_6(x)\ dx,\quad \left\{\color{red}{\text{apply integration by parts}}\right\}\\
&=2\sum_{n=1}^\infty\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(\frac{\zeta(6)}{n}-\frac{\zeta(5)}{n^2}+\frac{\zeta(4)}{n^3}-\frac{\zeta(3)}{n^4}+\frac{\zeta(2)}{n^5}-\frac{H_n}{n^6}\right)\\
3S_2+S_1&=2\sum_{n=1}^\infty\frac{H_n}{n^8}+2\zeta(6)\sum_{n=1}^\infty\frac{H_n}{n^2}-2\zeta(5)\sum_{n=1}^\infty\frac{H_n}{n^3}+2\zeta(4)\sum_{n=1}^\infty\frac{H_n}{n^4}\\
&\quad-2\zeta(3)\sum_{n=1}^\infty\frac{H_n}{n^5}+2\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n^6}-2\zeta(2)\zeta(7)\tag{1}
\end{align}
From Euler's identity, we can obtain the following results:
$$\sum_{n=1}^\infty\frac{H_n}{n^2}=2\zeta(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^3}=\frac54\zeta(4)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^5}=\frac74\zeta(6)-\frac12\zeta^2(3)$$
$$\sum_{n=1}^\infty\frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$
$$\sum_{k=1}^\infty\frac{H_k}{k^8}=5\zeta(9)-\zeta(2)\zeta(7)-\zeta(3)\zeta(6)-\zeta(4)\zeta(5)$$
By plugging these results along with that of $S_1$ in $(1)$, we get the closed form of $S_2$.