To calculate these two sums, we are going to establish two relations and solve them by elimination.
To establish the first relation, we use $\displaystyle I=\int_0^1\frac{\ln^4(1+x)+6\ln^2(1-x)\ln^2(1+x)}{x}\ dx=\frac{21}4\zeta(5)\tag{1}$
which was proved by Khalef Ruhemi ( unfortunately he is not an MSE user).
The proof as follows: using the algebraic identity $\ b^4+6a^2b^2=\frac12(a-b)^4+\frac12(a+b)^4-a^4$
with $\ a=\ln(1-x)$ and $\ b=\ln(1+x)$ , divide both sides by $x$ then integrate, we get
$$I=\frac12\underbrace{\int_0^1\frac1x{\ln^4\left(\frac{1-x}{1+x}\right)}\ dx}_{\frac{1-x}{1+x}=y}+\underbrace{\frac12\int_0^1\frac{\ln^4(1-x^2)}{x}\ dx}_{x^2=y}-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$
$$=\int_0^1\frac{\ln^4x}{1-x^2}+\frac14\int_0^1\frac{\ln^4(1-x)}{x}\ dx-\int_0^1\frac{\ln^4(1-x)}{x}\ dx$$
$$=\frac12\int_0^1\frac{\ln^4x}{1-x}+\frac12\int_0^1\frac{\ln^4x}{1+x}-\frac34\underbrace{\int_0^1\frac{\ln^4(1-x)}{x}\ dx}_{1-x=y}$$
$$=\frac12\int_0^1\frac{\ln^4x}{1+x}\ dx+\frac14\int_0^1\frac{\ln^4x}{1-x}\ dx=\frac12\left(\frac{45}{2}\zeta(5)\right)+\frac14(24\zeta(5))=\frac{21}4\zeta(5)$$
On the other hand, $\quad\displaystyle I=\underbrace{\int_0^1\frac{\ln^4(1+x)}{x}\ dx}_{I_1}+6\int_0^1\frac{\ln^2(1-x)\ln^2(1+x)}{x}\ dx$
Using $\ln^2(1+x)=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)x^n\ $ for the second integral, we get
\begin{align}
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\int_0^1x^{n-1}\ln^2(1-x)\ dx\\
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac{1}{n^2}\right)\left(\frac{H_n^2+H_n^{(2)}}{n}\right)\\
I&=I_1+12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^3+H_nH_n^{(2)}}{n^2}\right)-12\sum_{n=1}^\infty(-1)^n\left(\frac{H_n^2+H_n^{(2)}}{n^3}\right)\tag{2}
\end{align}
From $(1)$ and $(2)$, we get
$$\boxed{\small{R_1=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}+\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{7}{16}\zeta(5)+\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac{1}{12}I_1}}$$
and the first relation is established.
To get the second relation, we need to use the sterling number formula ( check here)
$$ \frac{\ln^k(1-x)}{k!}=\sum_{n=k}^\infty(-1)^k \begin{bmatrix} n \\ k \end{bmatrix}\frac{x^n}{n!}$$
letting $k=4$ and using $\displaystyle\begin{bmatrix} n \\ 4 \end{bmatrix}=\frac{1}{3!}(n-1)!\left[\left(H_{n-1}\right)^3-3H_{n-1}H_{n-1}^{(2)}+2H_{n-1}^{(3)}\right],$ we get $$\frac14\ln^4(1-x)=\sum_{n=1}^\infty\frac{x^{n+1}}{n+1}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
differentiate both sides with respect to $x$, we get
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
Now replace $x$ with $-x$ then multiply both sides by $\frac{\ln x}{x}$ and integrate, we get
$$-\sum_{n=1}^\infty(-1)^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)\int_0^1x^{n-1}\ln x\ dx=\int_0^1\frac{\ln^3(1+x)\ln x}{x(1+x)}\ dx$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx-\underbrace{\int_0^1\frac{\ln^3(1+x)\ln x}{1+x}\ dx}_{IBP}$$
$$\sum_{n=1}^\infty \frac{(-1)^n}{n^2}\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)=\int_0^1\frac{\ln^3(1+x)\ln x}{x}\ dx+\frac14I_1$$
Rearranging the terms, we get
$$\boxed{R_2=\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}-3\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\int_0^1\frac{\ln^3(1+x)\ln x}{x}-2\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}+\frac14I_1}$$
and the second relation is established.
Now we are ready to calculate the first sum.
\begin{align}
\sum_{n=1}^\infty\frac{(-1)^nH_n^3}{n^2}&=\frac{3R_1+R_2}{4}\\
&=\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac34\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}-\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}\\
&\quad+\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx+\frac{21}{64}\zeta(5)
\end{align}
the closed form of the first and second sum can be found here and the closed form of the third sum is evaluated here. as for the integral, I evaluated it here.
by combining these results, we get our closed form.
and the second sum.
$$\sum_{n=1}^\infty\frac{(-1)^nH_nH_n^{(2)}}{n^2}=\frac{R_1-R_2}{4}$$
$$\small{=\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{n^3}+\frac14\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n^3}+\frac12\sum_{n=1}^\infty\frac{(-1)^nH_n^{(3)}}{n^2}-\frac14\int_0^1\frac{\ln x\ln^3(1+x)}{x}\ dx-\frac1{12}I_1+\frac{7}{64}\zeta(5)}$$
lets calculate $I_1$ and by setting $\frac1{1+x}=y$, we get
\begin{align}
I_1&=\int_0^1\frac{\ln^4(1+x)}{x}=\int_{1/2}^1\frac{\ln^4x}{x}\ dx+\int_{1/2}^1\frac{\ln^4x}{1-x}\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty\int_{1/2}^1 x^{n-1}\ln^4x\ dx\\
&=\frac15\ln^52+\sum_{n=1}^\infty\left(\frac{24}{n^5}-\frac{24}{n^52^n}-\frac{24\ln2}{n^42^n}-\frac{12\ln^22}{n^32^n}-\frac{4\ln^32}{n^22^n}-\frac{\ln^42}{n2^n}\right)\\
&=4\ln^32\zeta(2)-\frac{21}2\ln^22\zeta(3)+24\zeta(5)-\frac45\ln^52-24\ln2\operatorname{Li}_4\left(\frac12\right)-24\operatorname{Li}_5\left(\frac12\right)
\end{align}
by combining the result of $I_1$ along with the results we used in our first sum, we get the closed form of the second sum.
UPDATE:
The identity used above:
$$-\frac{\ln^3(1-x)}{1-x}=\sum_{n=1}^\infty x^n\left(H_n^3-3H_nH_n^{(2)}+2H_n^{(3)}\right)$$
can also be proved this way.
Here is a sketch of Cornel's way.
We use that $\displaystyle \sum_{k=1}^{\infty} \frac{H_k}{(k+1)(k+2n+1)}=\frac{H_{2n}^2+H_{2n}^{(2)}}{4n}$ and then multiply all by $1/n^2$ and consider the summation from $n=1$ to $\infty$. Later in the process we use another critical step, $\displaystyle \sum_{n=1}^{\infty} \frac{1}{(2k+2n+1) 2n}=\frac{1}{(2k+1)^2}+\frac{H_{2k}}{2k+1}-\frac{H_k}{2(2k+1)}-\frac{\log(2)}{2k+1}$.
Essentially, these are almost exactly the steps presented in section 6.59, pages $530$-$532$, in the book (Almost) Impossible Integrals, Sums, and Series.
So, after simple calculations and rearrangements we arrive at
$$\frac{1}{4} \sum _{n=1}^{\infty } \frac{\left(H_{2 n}\right){}^2}{n^3}+\frac{1}{4} \sum _{n=1}^{\infty } \frac{H_{2 n}^{(2)}}{n^3}$$
$$=\frac{1}{8}\sum _{n=1}^{\infty }\frac{H_n}{n^4}-\frac{1}{4}\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}+\frac{\pi^2}{24}\sum _{n=1}^{\infty }\frac{H_{2 n}}{n^2}-4\sum _{n=1}^{\infty }\frac{H_{2 n+1}^2}{(2 n+1)^3}+\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^2}\\+4\sum _{n=1}^{\infty }\frac{H_{2 n+1}}{(2 n+1)^4}+2\sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{(2 n+1)^3}+4 \log (2)\sum _{n=1}^{\infty }\frac{ H_{2 n+1}}{(2 n+1)^3}-\frac{\pi^2}{48}\sum _{n=1}^{\infty }\frac{1}{n^3}-\frac{\pi^2}{6}\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^3}-4 \log (2)\sum _{n=1}^{\infty }\frac{1}{(2 n+1)^4}.$$
Since all the series are known except the desired one, the extraction is immediately achieved.
For example, a solution to the challenging series $\displaystyle \sum _{n=1}^{\infty }\frac{H_n H_{2 n}}{n^3}$ is presented in https://math.stackexchange.com/q/3345138.
The way to go also appears in details in the preprint On the calculation of two essential harmonic series with a weight $5$ structure, involving harmonic numbers of the type $H_{2n}$ by Cornel Ioan Valean
Best Answer
Expanding on my comment above:
Let $\mathcal{S}$ denote the value the following infinite series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}\frac{\overline{H}_{n}-H_{n/2}}{n^{3}}\approx0.260631,$$
where $\overline{H}_{n}$ here denotes the $n$-th alternating harmonic number and is defined for each positive integer $n$ by the finite series
$$\overline{H}_{n}:=\sum_{k=1}^{n}\frac{\left(-1\right)^{k-1}}{k};~~~\small{n\in\mathbb{N}},$$
and the $\alpha$-th harmonic number $H_{\alpha}$ is defined here for real argument $\alpha$ through Euler's integral representation
$$H_{\alpha}:=\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{\alpha}}{1-t};~~~\small{\alpha\in\left(-1,\infty\right)}.$$
An integral representation for the alternating harmonic numbers can be obtained as follows: for $n\in\mathbb{N}$, we have
$$\begin{align} \overline{H}_{n} &=\sum_{k=1}^{n}\frac{\left(-1\right)^{k-1}}{k}\\ &=\sum_{k=1}^{n}\left(-1\right)^{k-1}\int_{0}^{1}\mathrm{d}t\,t^{k-1}\\ &=\int_{0}^{1}\mathrm{d}t\,\sum_{k=1}^{n}\left(-t\right)^{k-1}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{1-\left(-t\right)^{n}}{1+t}.\\ \end{align}$$
The difference $\overline{H}_{n}-H_{n/2}$ can be expressed as a single definite integral by combining the representations above:
$$\begin{align} \overline{H}_{n}-H_{n/2} &=\int_{0}^{1}\mathrm{d}x\,\frac{1-\left(-x\right)^{n}}{1+x}-\int_{0}^{1}\mathrm{d}t\,\frac{1-t^{n/2}}{1-t}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1-\left(-x\right)^{n}}{1+x}-\int_{0}^{1}\mathrm{d}x\,\frac{2x\left(1-x^{n}\right)}{1-x^{2}};~~~\small{\left[t=x^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1-\left(-x\right)^{n}}{1+x}-\frac{2x\left(1-x^{n}\right)}{1-x^{2}}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{1-\left(-x\right)^{n}}{1+x}-\frac{1-x^{n}}{1-x}+\frac{1-x^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right].\\ \end{align}$$
Recall that for positive integer order $p$ and complex argument $z$, the $p$-th order polylogarithm $\operatorname{Li}_{p}{\left(z\right)}$ is defined on the unit disk by the infinite series
$$\operatorname{Li}_{p}{\left(z\right)}:=\sum_{n=1}^{\infty}\frac{z^{n}}{n^{p}};~~~\small{p>1\land\left|z\right|\le1}.$$
Using the technique of switching the order of summation and integration, it's then a straightforward matter to convert the series representation for $\mathcal{S}$ into a polylogarithmic integral. We find
$$\begin{align} \mathcal{S} &=\sum_{n=1}^{\infty}\frac{\overline{H}_{n}-H_{n/2}}{n^{3}}\\ &=\sum_{n=1}^{\infty}\frac{1}{n^{3}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\sum_{n=1}^{\infty}\frac{1}{n^{3}}\left[\frac{2}{1+x}-\frac{1-x^{n}}{1-x}-\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{2}{1+x}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{1-x^{n}}{1-x}-\sum_{n=1}^{\infty}\frac{1}{n^{3}}\cdot\frac{x^{n}+\left(-x\right)^{n}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{2\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}-\frac{\operatorname{Li}_{3}{\left(x\right)}+\operatorname{Li}_{3}{\left(-x\right)}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\right].\\ \end{align}$$
Define the function $F:\left(-\infty,1\right]\rightarrow\mathbb{R}$ via the polylogarithmic expression
$$F{\left(x\right)}:=\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right];~~~\small{x<1},$$
$$F{\left(1\right)}:=\lim_{x\to1^{-}}\bigg{[}\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\bigg{]}=\frac12\left[\operatorname{Li}_{2}{\left(1\right)}\right]^{2}.$$
Calculating the derivative of $F$, we find
$$\begin{align} \frac{d}{dx}F{\left(x\right)} &=\frac{d}{dx}\bigg{[}\frac12\left[\operatorname{Li}_{2}{\left(x\right)}\right]^{2}-\ln{\left(1-x\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\bigg{]}\\ &=-\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x}-\ln{\left(1-x\right)}\left[-\frac{\operatorname{Li}_{2}{\left(x\right)}}{x}\right]+\frac{1}{1-x}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}\right]\\ &=\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x},\\ \end{align}$$
i.e., $F{\left(x\right)}$ is an antiderivative of $\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}$, allowing us to reduce the integral form for $\mathcal{S}$ to
$$\begin{align} \mathcal{S} &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\right]\\ &=\int_{0}^{1}\mathrm{d}x\,\left[\frac{\operatorname{Li}_{3}{\left(1\right)}}{1+x}-\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(x\right)}}{1-x}+\frac{\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-x\right)}}{1+x}\right]-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{d}{dx}\left[\operatorname{Li}_{3}{\left(1\right)}\ln{\left(1+x\right)}-F{\left(x\right)}-F{\left(-x\right)}\right]-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\operatorname{Li}_{3}{\left(1\right)}\ln{\left(2\right)}-F{\left(1\right)}-F{\left(-1\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\operatorname{Li}_{3}{\left(1\right)}\ln{\left(2\right)}-\frac12\left[\operatorname{Li}_{2}{\left(1\right)}\right]^{2}-\frac12\left[\operatorname{Li}_{2}{\left(-1\right)}\right]^{2}+\ln{\left(2\right)}\left[\operatorname{Li}_{3}{\left(1\right)}-\operatorname{Li}_{3}{\left(-1\right)}\right]\\ &~~~~~-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\frac{11}{4}\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac58\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}.\\ \end{align}$$
It remains to evaluate the integral $\mathcal{I}:=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}$. Now, there is an extremely tidy way to calculate this integral in terms of Nielsen generalized polylogarithms:
$$\begin{align} \mathcal{I} &=\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{Li}_{3}{\left(x\right)}}{1+x}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1+x\right)}\operatorname{Li}_{2}{\left(x\right)}}{x};~~~\small{I.B.P.s}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\operatorname{Li}_{2}{\left(-x\right)}}{x};~~~\small{I.B.P.s}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x}\int_{0}^{1}\mathrm{d}y\,\frac{(-1)\ln{\left(1+xy\right)}}{y}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\,\frac{\ln{\left(1-x\right)}\ln{\left(1+xy\right)}}{xy}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\ln{\left(1+xy\right)}}{xy}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{1}{y}\int_{0}^{1}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}\ln{\left(1+yx\right)}}{x}\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{1}{y}\left[\operatorname{Li}_{3}{\left(-y\right)}+S_{1,2}{\left(-y\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\int_{0}^{1}\mathrm{d}y\,\frac{d}{dy}\left[\operatorname{Li}_{4}{\left(-y\right)}+S_{2,2}{\left(-y\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}-\left[\operatorname{Li}_{4}{\left(-1\right)}+S_{2,2}{\left(-1\right)}\right]\\ &=\ln{\left(2\right)}\,\zeta{\left(3\right)}-\frac12\left[\zeta{\left(2\right)}\right]^{2}+\frac78\zeta{\left(4\right)}-S_{2,2}{\left(-1\right)}.\\ \end{align}$$
To summarize, the evaluation of the series $\mathcal{S}$ can be boiled down to finding a closed-form expression for the Nielsen generalized polylogarithmic term $S_{2,2}{\left(-1\right)}$.