From here we have $$\arcsin^2z=\frac12\sum_{k=1}^\infty\frac{(2z)^{2k}}{k^2{2k \choose k}}$$
Set $z=i\sqrt{\frac{y}{8}}$, we get
$$-\text{arcsinh}^2\left(\sqrt{\frac{y}{8}}\right)=\frac12\sum_{k=1}^\infty\frac{(-1)^{k}y^k}{k^22^k{2k \choose k}}$$
Now multiply both sides by $\frac{2\ln y}{y}$ then integrate from $y=0$ to $1$, we get
\begin{align}
\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}&=-2\int_0^1\frac{\text{arcsinh}^2\left(\sqrt{\frac{y}{8}}\right)\ln y}{y}\ dy,\quad \color{red}{\text{arcsinh}\left(\sqrt{\frac{y}{8}}\right)=x}\\
&=-4\int_0^{\frac{\ln2}{2}}x^2\ln\left(8\sinh^2x\right)\coth x\ dx\\
&=-3\ln2\int_0^{\frac{\ln2}{2}}4x^2 \coth x\ dx-8\int_0^{\frac{\ln2}{2}}x^2\ln(\sinh x)\coth x\ dx\tag{1}
\end{align}
The first integral is calculated here
$$\int_0^{\frac{\ln2}{2}}4x^2 \coth x\ dx=\frac1{4}\zeta(3)-\frac1{6}\ln^32\tag{2}$$
As for the second integral, we compute it as follows
\begin{align}
I&=\int_0^{\frac{\ln2}{2}}x^2\ln(\sinh x)\coth x\ dx,\quad \color{red}{x=\ln y}\\
&=\int_0^{\sqrt{2}}\ln^2y\ln\left(\frac{y^2-1}{2y}\right)\left(\frac{y^2+1}{y^2-1}\right)\frac{\ dy}{y},\quad \color{red}{y^2-1=x}\\
&=\frac18\int_0^1\ln^2(1+x)\left(\ln x-\ln2-\frac12\ln(1+x)\right)\left(\frac{2}{x}-\frac{1}{1+x}\right)\ dx\\
&=\frac14\int_0^1\frac{\ln^2(1+x)\ln x}{x}\ dx-\frac14\ln 2\int_0^1\frac{\ln^2(1+x)}{x}\ dx-\frac18\int_0^1\frac{\ln^3(1+x)}{x}\ dx\\
&\quad-\frac18\underbrace{\int_0^1\frac{\ln^2(1+x)\ln x}{1+x}\ dx}_{\Large IBP}+\frac18\underbrace{\int_0^1\frac{\ln^2(1+x)}{1+x}\ dx}_{\Large\frac13\ln^32}+\frac1{16}\underbrace{\int_0^1\frac{\ln^3(1+x)}{1+x}\ dx}_{\Large\frac14\ln^42}\\
&=\frac14\underbrace{\int_0^1\frac{\ln^2(1+x)\ln x}{x}}_{\Large I_1}-\frac14\underbrace{\ln 2\int_0^1\frac{\ln^2(1+x)}{x}}_{\Large I_2}-\frac1{12}\underbrace{\int_0^1\frac{\ln^3(1+x)}{x}}_{\Large I_3}+\frac{11}{192}\ln^42\tag{3}
\end{align}
Lets start with the first one and by using :
$$\ln^2(1+x)=2\sum_{n=1}^\infty\frac{H_n}{n+1}(-x)^{n+1}=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)x^n$$
We get
\begin{align}
I_1&=\int_0^1\frac{\ln^2(1+x)\ln x}{x}\ dx=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)\int_0^1x^{n-1}\ln x\ dx\\
&=2\sum_{n=1}^\infty(-1)^n\left(\frac{H_n}{n}-\frac1{n^2}\right)\left(-\frac1{n^2}\right)\\
&=-2\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}+2\operatorname{Li}_4(-1)\\
&=-2\left(2\operatorname{Li_4}\left(\frac12\right)-\frac{11}4\zeta(4)+\frac74\ln2\zeta(3)-\frac12\ln^22\zeta(2)+\frac{1}{12}\ln^42\right)+2\left(-\frac78\zeta(4)\right)\\
&\boxed{=-4\operatorname{Li_4}\left(\frac12\right)+\frac{15}4\zeta(4)-\frac72\ln2\zeta(3)+\ln^22\zeta(2)-\frac{1}{6}\ln^42}
\end{align}
Note that $\sum_{n=1}^\infty(-1)^n\frac{H_n}{n^3}$ is calculated here.
\begin{align}
I_2&=\int_0^1\frac{\ln^2(1+x)}{x}\ dx\overset{x=\frac{1-y}{y}}{=}\int_{1/2}^1\frac{\ln^2x}{x(1-x)}\ dx\\
&=\int_{1/2}^1\frac{\ln^2x}{x}\ dx+\int_{1/2}^1\frac{\ln^2x}{1-x}\ dx\\
&=\frac13\ln^32+\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^2x \ dx\\
&=\frac13\ln^32+\sum_{n=1}^\infty\left(-\frac{\ln^22}{n2^n}-\frac{2\ln2}{n^22^n}-\frac{2}{n^32^n}+\frac{2}{n^3}\right)\\
&=\frac13\ln^32-\ln^32-2\ln2\operatorname{Li}_2\left(\frac12\right)-2\operatorname{Li}_3\left(\frac12\right)+2\zeta(3)\\
&\boxed{=\frac14\zeta(3)}
\end{align}
Similarly
\begin{align}
I_3&=\int_0^1\frac{\ln^3(1+x)}{x}\ dx\overset{x=\frac{1-y}{y}}{=}-\int_{1/2}^1\frac{\ln^3x}{x(1-x)}\ dx\\
&=-\int_{1/2}^1\frac{\ln^3x}{x}\ dx-\int_{1/2}^1\frac{\ln^3x}{1-x}\ dx\\
&=\frac14\ln^42-\sum_{n=1}^\infty\int_{1/2}^1x^{n-1}\ln^3x \ dx\\
&=\frac14\ln^42-\sum_{n=1}^\infty\left(\frac{\ln^32}{n2^n}+\frac{3\ln^22}{n^22^n}+\frac{6\ln2}{n^32^n}+\frac{6}{n^42^n}-\frac{6}{n^4}\right)\\
&=\frac14\ln^42-\ln^42-3\ln^22\operatorname{Li}_2\left(\frac12\right)-6\ln2\operatorname{Li}_3\left(\frac12\right)-6\operatorname{Li}_4\left(\frac12\right)+6\zeta(4)\\
&\boxed{=-6\operatorname{Li}_4\left(\frac12\right)+6\zeta(4)-\frac{21}{4}\ln2\zeta(3)+\frac32\ln^22\zeta(2)-\frac14\ln^42}
\end{align}
Note that for $I_2$ and $I_3$, we used
$$\operatorname{Li_2}\left( \frac12\right) =\frac12\zeta(2)-\frac12\ln^22$$
$$\operatorname{Li_3}\left( \frac12\right)=\frac78\zeta(3)-\frac12\ln2\zeta(2)+\frac16\ln^32$$
Plugging the boxed results of the integrals $I_1$, $I_2$ and $I_3$ in $(3)$, we get
$$I=-\frac12\operatorname{Li}_4\left(\frac12\right)+\frac7{16}\zeta(4)-\frac12\ln2\zeta(3)+\frac18\ln^22\zeta(2)+\frac{7}{192}\ln^42\tag{4}$$
Plugging the results from $(2)$ and $(4)$ in $(1)$, we get
$$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}=4\operatorname{Li}_4\left(\frac12\right)-\frac72\zeta(4)+\frac{13}4\ln2\zeta(3)-\ln^22\zeta(2)+\frac5{24}\ln^42$$
$$\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k^42^k{2k \choose k}}=-2\int_0^1\frac{\text{arcsinh}^2\left(\sqrt{\frac{y}{8}}\right)\ln y}{y}\ dy$$
– Varun Vejalla Jul 29 '19 at 02:52