In this solution, I proved
$$\sum_{n=1}^\infty \frac{H_n^{(3)}x^n}{n}=\operatorname{Li}_4(x)-\ln(1-x)\operatorname{Li}_3(x)-\frac12\operatorname{Li}^2_2(x)\tag{1}$$
Multiply both sides of $(1)$ by $\large \frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ and use the fact that $\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\overset{IBP}{=}\large
\frac{\zeta(2)}{n}-\frac{H_n}{n^2}$ to get
$$\small{\sum_{n=1}^\infty \frac{H_n^{(3)}}{n}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx-\underbrace{\int_0^1\frac{\ln(1-x)\operatorname{Li}_3(x)\operatorname{Li}_2(x)}{x}\ dx}_{IBP}-\frac12\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx}$$
$$\small{\zeta(2)\sum_{n=1}^\infty \frac{H_n^{(3)}}{n^2}-\sum_{n=1}^\infty \frac{H_nH_n^{(3)}}{n^3}=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx+\frac54\zeta(3)\zeta(4)-\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx}\tag{2}$$
By Cauchy product we have
$$\operatorname{Li}_2^2(x)=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)x^n\tag{3}$$
Divide both sides of $(3)$ by $x$ then integrate from $x=0$ to $1$ to get
$$\boxed{S=\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)\frac1n=\int_0^1\frac{\operatorname{Li}_2^2(x)}{x}\ dx}$$
Now multiply both sides of $(3)$ by $\large \frac{\operatorname{Li}_2(x)}{x}$ then integrate from $x=0$ to $1$ to get
$$\sum_{n=1}^\infty\left(\frac{4H_n}{n^3}+\frac{2H_n^{(2)}}{n^2}-\frac{6}{n^4}\right)\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx$$
$$\zeta(2)S-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}-2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+6\sum_{n=1}^\infty\frac{H_n}{n^6}=\int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx\tag{4}$$
By adding $(2)$ and $(4)$ and substituting the boxed value of $S=\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx$ we get
$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}\\=\small{-\frac54\zeta(3)\zeta(4)+6\sum_{n=1}^\infty\frac{H_n}{n^6}+\zeta(2)\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}+\zeta(2)\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx-\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx}$$
Now we are left with trivial integrals and lets start with the first one
\begin{align}
I_1&=\int_0^1\frac{\operatorname{Li}^2_2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^2}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\
&=\sum_{n=1}^\infty\frac1{n^2}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\zeta(2)\zeta(3)-\sum_{n=1}^\infty\frac{H_n}{n^4}
\end{align}
Similarly
\begin{align}
I_2&=\int_0^1\frac{\operatorname{Li}_4(x)\operatorname{Li}_2(x)}{x}\ dx=\sum_{n=1}^\infty\frac1{n^4}\int_0^1x^{n-1}\operatorname{Li}_2(x)\ dx\\
&=\sum_{n=1}^\infty\frac1{n^4}\left(\frac{\zeta(2)}{n}-\frac{H_n}{n^2}\right)=\zeta(2)\zeta(5)-\sum_{n=1}^\infty\frac{H_n}{n^6}
\end{align}
Combining $I_1$ and $I_2$ gives
$$2\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{n^4}+\sum_{n=1}^\infty\frac{H_nH_n^{(3)}}{n^3}\\=\frac54\zeta(3)\zeta(4)-\zeta(2)\zeta(5)-\zeta(2)\sum_{n=1}^\infty\frac{H_n}{n^4}+7\sum_{n=1}^\infty\frac{H_n}{n^6}+\zeta(2)\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-4\sum_{n=1}^\infty\frac{H_n^2}{n^5}$$
We have
$$S_1=\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$$
$$S_2=\sum_{n=1}^\infty\frac{H_n}{n^6}=4\zeta(7)-\zeta(2)\zeta(5)-\zeta(3)\zeta(4)$$
$$S_3=\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\frac{11}2\zeta(5)-2\zeta(2)\zeta(3)$$
$$S_4=\sum_{n=1}^\infty\frac{H_n^2}{n^5}=6\zeta(7)-\zeta(2)\zeta(5)-\frac52\zeta(3)\zeta(4)$$
By plugging the results of $S_1$, $S_2$, $S_3$ and $S_4$ we prove the equality of our problem.
Its interesting to see that the integral $\large \int_0^1\frac{\operatorname{Li}^3_2(x)}{x}\ dx$ got cancelled out which is really hard to crack.
Proofs: $S_1$ and $S_2$ can be found using Euler's identity, $S_4$ can be found here. As for $S_3$, we can calculate it as follows
Again,by Cauchy product we have
$$\operatorname{Li}_2(x)\operatorname{Li}_3(x)=\sum_{n=1}^\infty\left(\frac{6H_n}{n^4}+\frac{3H_n^{(2)}}{n^3}+\frac{H_n^{(3)}}{n^2}-\frac{10}{n^5}\right)x^n$$
set $x=1$ to get
$$\zeta(2)\zeta(3)=6\sum_{n=1}^\infty\frac{H_n}{n^4}+3\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}+\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}-10\zeta(5)\tag{5}$$
Now lets use the well-known identity
$$\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q}+\sum_{n=1}^\infty\frac{H_n^{(q)}}{n^p}=\zeta(p)\zeta(q)+\zeta(p+q)$$
set $p=2$ and $q=3$
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\zeta(2)\zeta(3)+\zeta(5)-\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}\tag{6}$$
Plugging $(6)$ in $(5)$ and rearrange the terms we get
$$\sum_{n=1}^\infty\frac{H_n^{(3)}}{n^2}=\zeta(2)\zeta(3)-\frac72\zeta(5)+3\sum_{n=1}^\infty\frac{H_n}{n^4}$$
Finally substitute $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ to get the closed form of $S_3.$ As a bonus, plug $S_3$ in $(6)$ to get
$$\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$$
BONUS:
Starting with the identity
$$\frac{\ln^2(1-x)}{1-x}=\sum_{n=1}^\infty (H_n^2-H_n^{(2)})x^n$$
multiply both sides by $\frac{\ln^2x}{2x}$ then integrate from $x=0$ to $1$ we get
\begin{align}
\sum_{n=1}^\infty\frac{H_n^2-H_n^{(2)}}{n^3}&=\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{x(1-x)}\ dx\\
&=\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx+\underbrace{\frac12\int_0^1\frac{\ln^2(1-x)\ln^2x}{1-x}\ dx}_{1-x\mapsto x}\\
&=\int_0^1\frac{\ln^2(1-x)\ln^2x}{x}\ dx=2\sum_{n=1}^\infty\frac{H_n}{n+1}\int_0^1 x^n \ln^2x\ dx\\
&=4\sum_{n=1}^\infty\frac{H_n}{(n+1)^4}=4\sum_{n=1}^\infty\frac{H_n}{n^4}-4\zeta(5)=\boxed{8\zeta(5)-4\zeta(2)\zeta(3)}
\end{align}
Thus $$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=8\zeta(5)-4\zeta(2)\zeta(3)+\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}\\=\frac72\zeta(5)-\zeta(2)\zeta(3)$$
