Let $g(a)=\frac{a}{1-a}, f(a)=\log(1-a)$ and $h(a)=\log^2(1-a)$. Let $\square^m$ denote a $m$-dimensional unit hypercube.

The following is an application or reformulation of the above infinite series evaluation.
$$\int_0^1\int_0^1\frac{h(\frac12xz)-xh(\frac12z)+xzh(\frac12)-zh(\frac12x)}{x(1-x)z(1-z)}dxdz=
\frac{\log^42}{12}-\frac{\pi^4}{144}+\zeta(2)\log^22+\zeta(3)\log2.$$
**Proof outline:** Since $\frac1k=\int_0^1x^{k-1}dx$, we may write
\begin{align} \sum_{n\geq1}\frac{t^nH_n^3}{n+1}&=\sum_{n\geq1}t^n\int_0^1w^ndw\sum_{i,j,k=1}^n\int_{\square^3}x^{i-1}y^{j-1}z^{k-1}dxdydz \\
&=\int_{\square^4}dxdydzdw\sum_{n\geq1}(tw)^n\sum_{i,j,k=1}^nx^{i-1}y^{j-1}z^{k-1}\\
&=\int_{\square^4}\frac{dxdydzdw}{(1-x)(1-y)(1-z)}\sum_{n\geq1}(tw)^n(1-x^n)(1-y^n)(1-z^n)\\
&=\int_{\square^4}dxdydzdq\frac{g(tw)-g(twy)-g(twx)+g(twxy)-g(twz)+g(twyz)+g(twxz)-g(twxyz)}{(1-x)(1-y)(1-z)}.
\end{align}
Next, make a repeated use of the integral evaluations
$$\int_0^1\frac{g(tw)-g(twy)}{1-y}dy=-\frac{f(tw)}{1-tw}$$
and
$$\int_0^1\left(\frac{f(twx)}{1-twx}-\frac{f(tw)}{1-tw}\right)\frac{dw}{1-w}=
\frac{xh(t)-h(tx)}{2tx}$$
to arrival at the double integral upon replacing $t=\frac12$. The rest follows from the series evaluation algorithm that Julian Rosen alluded to. $\square$

**Remark.** This approach actually could prove (directly) the easier-looking
$$\sum_{n\geq1}\frac{H_n^2}{(n+1)2^n}=\frac13\log^32+\zeta(2)\log2-\frac12\zeta(3).$$