First, in view of Legrende's duplication formula,
$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}=2\sqrt{\pi}\sum_{n=1}^\infty\frac{\Gamma(2n)}{\Gamma(n)\,n!\,(2n+1)^4\, 16^n}
\\=-\frac{\sqrt{\pi}}{3}\int_0^1 \ln^3(x)\sum_{n=1}^{\infty}\frac{\Gamma(2n)}{\Gamma(n)\,n!}\left(\frac{x^2}{16}\right)^ndx\\
=-\sqrt{\pi}-\frac{\sqrt{\pi}}{6}\int_0^1\frac{\ln^3(x)}{\sqrt{1-x^2/4}}dx=-\sqrt{\pi}-\frac{\sqrt{\pi}}{3}\int_0^{\frac{\pi}{6}}\ln^3(2\sin x)dx$$
Claim: for $0<a\leq \frac{\pi}{2}$,
$$\int_0^a \ln^3\left(\frac{\sin x}{\sin a}\right)dx\tag{0}=\frac{4a-3\pi}{2}a^2\ln(2\sin a)-\frac{3\pi}{4}\zeta(3)+3\left(\frac{\pi}{2}-a\right)\Re\left(\frac12 \operatorname{Li}_3(e^{2ia})+\operatorname{Li}_3(1-e^{2ia})\right)+3\Im\left(\frac14\operatorname{Li}_4(e^{2ia})+\operatorname{Li}_4(1-e^{2ia})\right)
$$
Proof.
The idea is exactly identical to the proof displayed in this question. The proof is rather tedious (and obviously inefficient), and ends with a somewhat of a cancellation (implying the existence of a shortcut) , so I omit the boring algebra and outline the main ideas, which can be repeated systematically to obtain closed forms for even higher powers of logsine.
things to know:
$$\ln(2\sin x)=\ln(1-e^{2ix})+i\left(\frac{\pi}{2}-x\right) \tag{1}$$
$$\small\int\frac{\ln^3(1-x)}{x}dx=\ln^3(1-x)\ln(x)+3\ln^2(1-x)\text{Li}_2(1-x)-6\ln(1-x)\text{Li}_3(1-x)+6\text{Li}_4(1-x) \tag{2}$$
$$\int_0^a x\ln(2\sin x)dx=-\frac{a}{2}\text{Cl}_2(2a)-\frac14\Re\text{Li}_3(e^{2ia})+\frac{\zeta(3)}{4}\tag{3}$$
$$\int_0^a x^2\ln(2\sin x)dx=-\frac{a^2}{2}\text{Cl}_2(2a)-\frac{a}{2}\Re\text{Li}_3(e^{2ia})+\frac14\Im\text{Li}_4(e^{2ia})\tag{4}$$
$$\int_0^a \ln(\sin x)dx=-a\ln2-\frac12 \text{Cl}_2(2a)\tag{5}$$
$$\int_0^a \ln^2(\sin x)dx=\frac{a^3}{3}+a\ln^2 2-a\ln^2(2\sin a)-\ln(\sin a)\text{Cl}_2(2a)-\Im\text{Li}_3(1-e^{2ia})\tag{6}$$
$(1)$ is trivial, $(2)$ is not too hard to find, $(5)$ and $(6)$ are shown in the linked answer, and $(3)$&$(4)$ are easily found using $\,\,\ln(2\sin x)=-\sum_{n\geq1}\frac{\cos(2xn)}{n}$.
It is obvious that since we have $(5)$ and $(6)$, the claim $(0)$ depends on a closed form for $\displaystyle\int_0^a \ln^3(\sin x)dx$, and the latter may be evaluated in terms of $\displaystyle\int_0^a \ln^3(2\sin x)dx$.
But, with the help of $(1)$,
$$\int_0^a \ln^3(2\sin x)dx=\Re\int_0^a \ln^3(1-e^{2ix})dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx\\
=\frac12\Im\int_1^{e^{2ia}}\frac{\ln^3(1-x)}{x} dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx$$
(Same idea @RandomVariable had in this answer.)
Now we employ $(2),(3),(4),$ and $(5)$. Some expressions cancel and claim follows.$\square $
This result, together with the fact that $e^{i\pi/3}$ and $1-e^{i\pi/3}$ are conjugates, yields $\displaystyle \int_0^{\frac{\pi}{6}} \ln^3(2\sin x)dx=-\frac{\pi}{4}\zeta(3)-\frac94\Im\text{Li}_4(e^{i\pi/3})$,
and
$$S=\sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{9}{12}\Im\text{Li}_4(e^{i\pi/3})-1\right)$$
This form is equivalent to @user153012's form, as
$$\frac{2}{\sqrt{3}}\Im\text{Li}_4(e^{i\pi/3})=\sum_{n\geq 0}\frac{(-1)^n}{(3n+1)^4}+\sum_{n\geq 0}\frac{(-1)^n}{(3n+2)^4} \\=\frac{\psi^{(3)}\left(\frac13\right)}{216}-\frac{\pi^4}{81}$$
Also, as noted in the comments in the linked question, this may be used to write a closed form for a certain hypergeometric function.
This serves as a generalisation for the series, because $\displaystyle \sum_{n=1}^{\infty} \frac{\Gamma(n+1/2)}{(2n+1)^4 n!}a^{2n}=-\sqrt{\pi}\left(1+\frac1{6a}\int_0^{\sin^{-1} a}\ln^3\left(\frac{\sin x}{a}\right)dx\right)$
As an example, using closed forms for trilogarithms displayed in this post, we have
$$\int_0^{\frac{\pi}{4}}\ln^3(\sqrt{2}\sin x)dx=-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)$$
where $\beta(4)=\Im\text{Li}_4(i)$ is a value of Dirichlet's beta function.
Or equivalently,
$$\sum_{n=1}^{\infty} \frac{\Gamma\left(n+\frac12\right)}{(2n+1)^4\,2^n\,n!}=-\sqrt{\pi}-\frac{\sqrt{2\pi}}{6}\left(-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)\right)$$
Best Answer
A simpler way to evaluate it is to use the following identity: $$\boxed{\int _0^1\frac{x\ln ^n\left(t\right)}{1-xt}\:dt=\left(-1\right)^nn!\operatorname{Li}_{n+1}\left(x\right)}$$
Applying it to the desired integral yields:
$$\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=\int _0^1\ln \left(t\right)\int _0^1\frac{1-x}{1+x\left(1-x\right)t}\:dx\:dt$$ $$=-2\int _0^1\frac{\ln \left(t\right)\arctan \left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{\sqrt{-4-t}\sqrt{t}}dt=2\int _0^1\frac{\arctan ^2\left(\frac{\sqrt{t}}{\sqrt{-4-t}}\right)}{t}\:dt=-2\int _0^1\frac{\operatorname{arctanh} ^2\left(\sqrt{\frac{t}{4+t}}\right)}{t}\:dt$$ $$=-4\int _0^{\frac{1}{\sqrt{5}}}\frac{\operatorname{arctanh} ^2\left(x\right)}{x\left(1-x^2\right)}\:dx=-\frac{1}{2}\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{x}\:dx-\int _{\frac{1}{\phi ^2}}^1\frac{\ln ^2\left(x\right)}{1-x}\:dx$$ $$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\sum _{k=1}^{\infty }\frac{1}{k^3}+2\sum _{k=1}^{\infty }\frac{1}{k^3\:\phi ^{2k}}+4\ln \left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k^2\:\phi ^{2k}}+4\ln ^2\left(\phi \right)\sum _{k=1}^{\infty }\frac{1}{k\:\phi ^{2k}}$$ $$=\frac{8}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+2\operatorname{Li}_3\left(\frac{1}{\phi ^2}\right)+4\ln \left(\phi \right)\operatorname{Li}_2\left(\frac{1}{\phi ^2}\right)$$ By the use of standard functional equations we have: $$=-\frac{4}{3}\ln ^3\left(\phi \right)-2\zeta \left(3\right)+\frac{8}{5}\zeta \left(3\right)+\frac{4}{3}\ln ^3\left(\phi \right)-\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)+\frac{8}{5}\ln \left(\phi \right)\zeta \left(2\right)$$
And thus: $$\boxed{\int _0^1\frac{\operatorname{Li}_2\left(x\left(x-1\right)\right)}{x}\:dx=-\frac{2}{5}\zeta \left(3\right)}$$