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,
$$\sum_{n=1}^{\infty} \frac{\Gamma\left(n+\frac12\right)}{(2n+1)^4\,2^n\,n!}=-\sqrt{\pi}-\frac{\sqrt{2\pi}}{6}\left(-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)\right)$$
A short proof.
$\displaystyle Q_0(x) :=\Gamma(x+1)=\lim_{n\to\infty}\frac{n^x}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)}~~ , ~~~~ Q_1(x) :=\lim_{n\to\infty}\frac{e^{xn}n^{-x^2/2}}{\prod\limits_{k=1}^n\left(1+\frac{x}{k}\right)^k}$
$\displaystyle \pi^{1/2} = Q_0\left(-\frac{1}{2}\right)~~ , ~~ A^3 = 2^{7/12}Q_1\left(-\frac{1}{2}\right)^2 ~~ , ~~ e^{G/\pi} = 2^{3/4}\left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
It follows:
$\displaystyle \prod\limits_{k=1}^{2n}\left(1-\frac{1}{2k}\right)^{(-1)^k k} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{4k}\right)^{2k}}{\left(1-\frac{1}{4k-2}\right)^{2k-1}} = \prod\limits_{k=1}^n \frac{\left(1-\frac{1}{2k}\right)^{2k-1} \left(1-\frac{1}{4k}\right)^{2k}}{ \left(1-\frac{3}{4k}\right)^{2k-1} } = $
$\displaystyle = \frac{ \prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right) n^{-1/2} }{ n^{-3/4} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right) } \left(\frac{ e^{-3n/4} n^{-9/32} \prod\limits_{k=1}^n \left(1-\frac{1}{2k}\right)^k \prod\limits_{k=1}^n \left(1-\frac{1}{4k}\right)^k }{\prod\limits_{k=1}^n \left(1-\frac{3}{4k}\right)^k e^{-n/2} n^{-1/8} e^{-n/4} n^{-1/32} }\right)^2$
$\displaystyle \to ~\frac{ Q_0\left(-\frac{1}{2}\right) }{ Q_0\left(-\frac{3}{4}\right) } \left(\frac{ Q_1\left(-\frac{3}{4}\right) }{ Q_1\left(-\frac{1}{2}\right) Q_1\left(-\frac{1}{4}\right) }\right)^2$
$\displaystyle = 2^{-1/6} \frac{ Q_0\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{1}{2}\right) } \cdot 2^{-7/12} Q_1\left(-\frac{1}{2}\right)^{-2} \cdot 2^{3/4} \left(\frac{ Q_0\left(-\frac{1}{2}\right) Q_1\left(-\frac{3}{4}\right) }{ Q_0\left(-\frac{3}{4}\right) Q_1\left(-\frac{1}{4}\right)}\right)^2$
$\displaystyle = \Gamma\left(\frac{1}{4}\right) 2^{-1/6} \pi^{-1/2} A^{-3} e^{G/\pi}$
Best Answer
The main thing to see is that it has some kind of closed-form for $s\ge 2$ even and $q\ne 0$.
This is because you want to evaluate $\frac{g(0)-q^{-s}}{2}$ where $$g(z)=\sum_{k\in \Bbb{Z}} \frac{(-1)^k}{(z+k)^s+q^s}=\sum_{k\in \Bbb{Z}} (-1)^k\sum_{m=1}^s \frac{1}{c_m (z+k-a_m)}=\sum_{m=1}^s \frac{\pi}{c_m \sin(\pi (z-a_m))}$$ $$ a_m=qe^{i\pi (2m+1)/s}, c_m= sa_m^{s-1}$$
It works the same way for $s$ odd except that $\sum_{k\ge 1} \frac{(-1)^k}{(z+k)^s+q^s}$ isn't periodic anymore, instead of $\pi/\sin(\pi z)$ you'll have $\Gamma'/\Gamma(z-1)-\Gamma'/\Gamma(z/2-1)$ which does have only a few closed-forms (for $z$ rational where it can be expressed as a linear combination of $\log e^{2i\pi l /r}$).
It is supposedly obvious that there is no closed-form for $s\not\in \Bbb{Z}$.