Let $\mathcal{S}$ denote the sum of the following infinite series:
$$\mathcal{S}:=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}},$$
where here $\overline{H}_{n}$ denotes the $n$-th skew-harmonic number, defined for each positive integer $n$ by the finite sum
$$\overline{H}_{n}:=\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k};~~~\small{n\in\mathbb{N}}.$$
As you point out, it can be shown that the skew-harmonic numbers have the following generating function
$$\sum_{n=1}^{\infty}z^{n}\,\overline{H}_{n}=\frac{\ln{\left(1+z\right)}}{1-z};~~~\small{|z|<1}.$$
Using this, as well as the fact that
$$n\binom{2n}{n}\operatorname{B}{\left(n,n\right)}=2;~~~\small{n\in\mathbb{N}},$$
we find
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\
&=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\operatorname{B}{\left(n,n\right)}\\
&=\frac12\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}\int_{0}^{1}\mathrm{d}t\,t^{n-1}\left(1-t\right)^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}(-1)^{n-1}\overline{H}_{n}t^{n-1}\left(1-t\right)^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n-1}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{1}{\left[-t\left(1-t\right)\right]}\sum_{n=1}^{\infty}\overline{H}_{n}\left[-t\left(1-t\right)\right]^{n}\\
&=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{(-1)}{t\left(1-t\right)}\cdot\frac{\ln{\left(1-t(1-t)\right)}}{1+t(1-t)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t\left(1-t\right)\left(1+t-t^{2}\right)}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1-t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}-\frac12\int_{0}^{1}\mathrm{d}u\,\frac{\ln{\left(1-u+u^{2}\right)}}{u};~~~\small{\left[t=1-u\right]}\\
&~~~~~+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}.\\
\end{align}$$
Consider the following two-variable variant of the dilogarithm:
$$\operatorname{Li}_{2}{\left(r,\theta\right)}:=-\frac12\int_{0}^{r}\mathrm{d}t\,\frac{\ln{\left(1-2t\cos{\left(\theta\right)}+t^{2}\right)}}{t};~~~\small{\left(r,\theta\right)\in\mathbb{R}^{2}}.$$
It can be shown that
$$\operatorname{Li}_{2}{\left(1,\theta\right)}=\frac14\left(\pi-\theta\right)^{2}-\frac{\pi^{2}}{12};~~~\small{0<\theta<\pi}.$$
The following particular case can be used to evaluate the first integral of $\mathcal{S}$:
$$-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}=2\operatorname{Li}_{2}{\left(1,\frac{\pi}{3}\right)}=\frac12\left(\pi-\frac{\pi}{3}\right)^{2}-\frac{\pi^{2}}{6}=\frac{\pi^{2}}{18}.$$
Let $\mathcal{I}$ denote the value of the remaining definite integral:
$$\mathcal{I}:=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\approx0.0776351.$$
We find that $\mathcal{I}$ can be expressed in terms of the two-variable dilogarithm as follows:
$$\begin{align}
\mathcal{I}
&=-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\frac12\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}-\frac12\int_{\frac12}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}};~~~\small{symmetry}\\
&=-\int_{0}^{\frac12}\mathrm{d}t\,\frac{\ln{\left(\frac34+\left(\frac12-t\right)^{2}\right)}}{\frac54-\left(\frac12-t\right)^{2}}\\
&=-\int_{0}^{\frac12}\mathrm{d}u\,\frac{\ln{\left(\frac34+u^{2}\right)}}{\frac54-u^{2}};~~~\small{\left[t=\frac12-u\right]}\\
&=-\frac{2}{\sqrt{5}}\int_{0}^{\frac{1}{\sqrt{5}}}\mathrm{d}v\,\frac{\ln{\left(\frac34+\frac54v^{2}\right)}}{1-v^{2}};~~~\small{\left[u=\frac{\sqrt{5}}{2}v\right]}\\
&=-\frac{1}{\sqrt{5}}\int_{1}^{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}\mathrm{d}w\,\frac{(-2)}{(1+w)^{2}}\cdot\frac{(1+w)^{2}}{2w}\ln{\left(\frac{3+5\left(\frac{1-w}{1+w}\right)^{2}}{4}\right)};~~~\small{\left[v=\frac{1-w}{1+w}\right]}\\
&=-\frac{1}{\sqrt{5}}\int_{\frac{1-\frac{1}{\sqrt{5}}}{1+\frac{1}{\sqrt{5}}}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{3(1+w)^{2}+5(1-w)^{2}}{4(1+w)^{2}}\right)}\\
&=-\frac{1}{\sqrt{5}}\int_{\frac{\sqrt{5}-1}{\sqrt{5}+1}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)}\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\ln{\left(\frac{2-w+2w^{2}}{(1+w)^{2}}\right)};~~~\small{\phi:=\frac{1+\sqrt{5}}{2}}\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{1}{w}\left[\ln{\left(2\right)}-2\ln{\left(1+w\right)}+\ln{\left(1-\frac12w+w^{2}\right)}\right]\\
&=-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(2\right)}}{w}+\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{2\ln{\left(1+w\right)}}{w}\\
&~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-\frac12w+w^{2}\right)}}{w}\\
&=-\frac{\ln{\left(2\right)}}{\sqrt{5}}\ln{\left(\phi^{2}\right)}-\frac{2}{\sqrt{5}}\int_{-1}^{-\phi^{-2}}\mathrm{d}x\,\frac{\ln{\left(1-x\right)}}{x};~~~\small{\left[w=-x\right]}\\
&~~~~~-\frac{1}{\sqrt{5}}\int_{\phi^{-2}}^{1}\mathrm{d}w\,\frac{\ln{\left(1-2w\cos{\left(\theta\right)}+w^{2}\right)}}{w};~~~\small{\left[\theta:=\frac{\pi}{2}-\arcsin{\left(\frac14\right)}\right]}\\
&=-\frac{2\ln{\left(2\right)}\ln{\left(\phi\right)}}{\sqrt{5}}+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(-1\right)}\right]\\
&~~~~~+\frac{2}{\sqrt{5}}\left[\operatorname{Li}_{2}{\left(1,\theta\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\
&=\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right].\\
\end{align}$$
Hence,
$$\begin{align}
\mathcal{S}
&=\sum_{n=1}^{\infty}(-1)^{n-1}\frac{\overline{H}_{n}}{n\binom{2n}{n}}\\
&=-\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{t}+\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\ln{\left(1-t+t^{2}\right)}}{1+t-t^{2}}\\
&=\frac{\pi^{2}}{18}-\mathcal{I}\\
&=\frac{\pi^{2}}{18}-\frac{2}{\sqrt{5}}\left[-\ln{\left(2\right)}\ln{\left(\phi\right)}+\frac14\left(\pi-\theta\right)^{2}+\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}-\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}\right]\\
&=\frac{\pi^{2}}{18}+\frac{2}{\sqrt{5}}\ln{\left(2\right)}\ln{\left(\phi\right)}-\frac{1}{2\sqrt{5}}\left(\pi-\theta\right)^{2}-\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(-\phi^{-2}\right)}+\frac{2}{\sqrt{5}}\operatorname{Li}_{2}{\left(\phi^{-2},\theta\right)}.\\
\end{align}$$
Best Answer
With $S$ the constant to be computed, we have the following explicit yellow expression: $$ \begin{aligned} S&=\sum\frac{(-1)^{n-1}}{4^n\: n^2}\binom{2n}n\\ &=2\int_0^{-1}(\log2 -\log(1+\sqrt{1-y}))\;\frac{dy}y\qquad\text{ (OP formula)} \\ &=2\int_0^1\log\underbrace{\left(\frac{1+\sqrt{1+y}}2\right)}_{=:t}\;\frac{dy}y \\ & =2\int_1^{(1+\sqrt 2)/2=:a}\frac{2t-1}{t(t-1)}\;\log t\;dt = 2\int_1^a\left(\frac{t}{t(t-1)}+\frac{t-1}{t(t-1)}\right) \;\log t\;dt \\ &= 2\int_1^a\frac{\log t}{t-1}\; dt+ 2\int_1^a\frac{\log t}t\; dt \\ &= -2\Bigg[\operatorname{Li}_2(1-t)\Bigg]_1^a+ \Bigg[\log^2 t\Bigg]_1^a \\ &=-2\operatorname{Li}_2(1-a)+\log^2 a =\bbox[yellow]{-2\operatorname{Li}_2\left(\frac{1-\sqrt2}2\right)+\log^2\left(\frac{1+\sqrt2}2\right)}\ . \end{aligned} $$
Numerical checks, here pari/gp