Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$

Let's continue from where you showed that: $$S=\int_0^1\int_0^1\frac{1}{(1+x)(1+y)}\left(\frac{\ln(1-xy)-\ln2}{1+xy}\right)dydx$$ $$\overset{xy=t}=\int_0^1\int_0^x \frac{1}{(1+x)(x+t)}\frac{\ln\left(\frac{1-t}{2}\right)}{1+t}dtdx=\int_0^1\color{blue}{\int_t^1\frac{1}{(1+x)(x+t)}}\frac{\ln\left(\frac{1-t}{2}\right)}{1+t}\color{blue}{dx}dt $$ $$=\int_0^1 \frac{\color{blue}{\ln\left(\frac{(1+t)^2}{4t}\right)}\ln\left(\frac{1-t}{2}\right)}{\color{blue}{(1-t)}(1+t)}dt\overset{\large t\to\frac{1-x}{1+x}}=\frac12 \int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x}{x}\right)}{x}dx$$ $$=\frac12\int_0^1 \frac{\ln(1-x^2)\ln(1+x)}{x}dx-\frac12\int_0^1 \frac{\ln(1-x^2)\ln x}{x}dx$$ The first integral equals $-\frac{3\zeta(3)}{8}$ and can be found by plugging $m,n,q=1$ and $p=0$ in this general result: $$\small \int_0^1 \frac{[m\ln(1+x)+n\ln(1-x)][q\ln(1+x)+p\ln(1-x)]}{x}dx=\left(\frac{mq}{4}-\frac{5}{8}(mp+nq)+2np\right)\zeta(3)$$ The second integral can be expanded into power series: $$\int_0^1 \frac{\ln(1-x^2)\ln x}{x}dx\overset{x^2=t}=\frac14\int_0^1 \frac{\ln(1-t)\ln t}{t}dt$$ $$=-\frac14\sum_{n=1}^\infty \frac{1}{n}\int_0^1t^{n-1}\ln t\, dt=\frac14\sum_{n=1}^\infty \frac{1}{n^3}=\frac{\zeta(3)}{4}$$ $$\Rightarrow \sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=\frac12\left(-\frac{3\zeta(3)}{8}-\frac{\zeta(3)}{4}\right)=-\frac{5\zeta(3)}{16}$$

Different way to calculate $\mathcal{I}_1-\mathcal{I}_2$:

We have from the previous solution

$$\mathcal{I}_1=\int_0^1\ln x\frac{\ln(1+x)-\ln x}{1+x}dx$$

$$=\underbrace{\int_0^1\frac{\ln x\ln(1+x)}{1+x}dx}_{IBP}-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=-\frac12\int_0^1\frac{\ln^2(1+x)}{x}dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

and

$$\mathcal{I}_2=\int_0^1\frac{\tanh^{-1}x\ln(1-x)}{x}dx$$

$$=-\frac12\int_0^1\frac{\ln^2(1-x)}{x}dx+\frac12\int_0^1\frac{\ln(1-x)\ln(1+x)}{x}dx$$

so

$$\mathcal{I}_1-\mathcal{I}_2=\frac12\int_0^1\frac{\color{blue}{\ln^2(1-x)-\ln^2(1+x)-\ln(1-x)\ln(1+x)}}{x}\ dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

now use the algebraic identity

$$a^2-b^2-ab=2a^2-\frac34(a+b)^2-\frac14(a-b)^2$$

with $a=\ln(1-x)$ and $b=\ln(1+x)$ we get

$$\mathcal{I}_1-\mathcal{I}_2=\frac12\int_0^1\frac{\color{blue}{\ln^2(1-x)-\frac34\ln^2(1-x^2)-\frac14\ln^2\left(\frac{1-x}{1+x}\right)}}{x}\ dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\int_0^1\frac{\ln^2(1-x)}{x}dx-\frac38\underbrace{\int_0^1\frac{\ln^2(1-x^2)}{x}dx}_{\large x^2\mapsto x}-\frac18\underbrace{\int_0^1\frac{\ln^2\left(\frac{1-x}{1+x}\right)}{x}dx}_{\large\frac{1-x}{1+x}\mapsto x}-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\frac{13}{16}\underbrace{\int_0^1\frac{\ln^2(1-x)}{x}dx}_{1-x\mapsto x}-\frac14\int_0^1\frac{\ln^2x}{1-x^2}dx-\int_0^1\frac{\ln^2x}{1+x}dx$$

$$=\frac{11}{16}\int_0^1\frac{\ln^2x}{1-x}dx-\frac98\int_0^1\frac{\ln^2x}{1+x}dx=-\frac{5}{16}\zeta(3)$$