Thank you for posting this question, I enjoyed trying to answer it.
Start with the expression that Mathematica gave you and replace each
argument $\frac14$ of a hypergeometric function with $\frac z4$,
because we will be taking limits. I will call the two hypergeometric
functions $Q_1(z)$ and $Q_2(z)$. Each term can be brought to a closed
form by using identity 16.6.2 from the DLMF.
Setting $a=\frac12$, $b=1-\frac\nu2$, we get
$$ Q_1(z) = (1-z)^{-\frac12} F\left(\frac16,
\frac36,\frac 56; 1-\frac\nu2, 1+\frac\nu2; \frac{-27 z}{4(1-z)^3}
\right), $$
and setting $a=\frac{1+3\nu}{2}$, $b=1+\frac\nu2$, we get
$$ Q_2(z) = (1-z)^{-\frac{1+3\nu}{2}} F\left(\frac{1+3\nu}6,
\frac{3+3\nu}{6}, \frac{5+3\nu}{6}; 1+\frac\nu2, 1+\nu;
\frac{-27z}{4(1-z)^3} \right). $$
(Note that there are 6 possible identities to try per function, one for each possible choice of
$a$ and $b$ from the parameters, so it helps to do this on a
computer.)
The reason this works is that now the point $z=1$ is a singular point
of the hypergeometric functions on the right hand side, and Mathematica
will succeed in finding the limits as $z\to1$. The expression for the
whole integral that you have is
$$ Q =
\frac{2^{\frac43}\pi^{\frac12}}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)\Gamma(\frac56-\frac\nu2)\sin\frac{\nu\pi}2}
\left( -1 + 3^{-\frac{3\nu}2}\cos\left(\frac{\nu\pi}{2}\right)
\frac{\Gamma(\frac{1+3\nu}{2})
\Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}
\right). $$
Call the large expression in brackets $A$, and then write
$$ A = -1 + B 3^{-\frac{3\nu}{2}}\cos\frac{\pi\nu}{2}, \qquad B = \frac{\Gamma(\frac{1+3\nu}{2})
\Gamma(\frac56-\frac\nu2)}{\Gamma(\frac{1+\nu}2)\Gamma(\frac56+\frac\nu2)}. $$
Now, Mathematica will not simplify $A$ or $B$ on its own, so it needs
help. Set $x=\frac16+\frac\nu2$, and use the multiplication formula to
get
$$
\frac{\Gamma(\frac{1+3\nu}2)}{\Gamma(\frac{1+\nu}{2})\Gamma(\frac56+\frac\nu2)}
= \frac{\Gamma(3x)}{\Gamma(x+\frac13)\Gamma(x+\frac23)} =
\frac{\Gamma(x)}{2\pi} 3^{3x-1/2}. $$
After this, $A$ simplifies to
$$ A = -1 +
\frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{2\pi}\cos\frac{\pi\nu}{2}
= -1 + \frac{\cos\frac{\pi\nu}{2}}{2\sin(\frac\pi6+\frac{\pi\nu}{2})},
$$
where I've also used the reflection formula for $\Gamma(z)\Gamma(1-z)$ to get rid of the
gamma functions. Some further amount of manual trigonometry yields
$$ A = -\frac{\sqrt{3}}{2}\frac{\sin\frac{\pi\nu}{2}}{\sin(\frac\pi6 +
\frac{\nu\pi}{2})}. $$
Finally, write
$$ \frac{1}{\sin(\frac\pi6+\frac{\pi\nu}{2})} =
\frac{\Gamma(\frac16+\frac\nu2)\Gamma(\frac56-\frac\nu2)}{\pi}, $$
and substitute back. Lots of things cancel, and the answer is
$$ Q = -\frac{3^{1/2}2^{1/3}}{\pi^{1/2}}
\frac{\Gamma(\frac16+\frac\nu2)}{\Gamma(-\frac16)\Gamma(\frac56+\frac\nu2)}. $$
This closed form is equivalent to the one you gave through the
use of $\Gamma(\frac16)\Gamma(-\frac16)=-12\pi$.
P.S.
I would also like to note that the integral
$$ I(\nu,c) = \int_0^\infty J_\nu(x)^2 J_\nu(c x)\,dx $$
and its general form
$$ \int_0^\infty x^{\rho-1}J_\nu(a x) J_\mu(b x) J_\lambda(c x)\,dx $$
appear in Gradshteyn and Ryzhik, and you can find a paper "Some
infinite integrals involving bessel functions, I and II" by
W. N. Bailey, which evaluates this integral in terms of Appell
functions, but only in the case $c>2$ ($|c|>|a|+|b|$), which is where the $F_4$ Appell
function converges. DLMF 16.16.6 actually gives a way to write this
integral as
$$
\frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(1+\nu)^2\Gamma(\frac{1-\nu}{2})}
\,\,\,{}_2F_1\left( \frac{1+\nu}{2}, \frac{1+3\nu}{2}; 1+\nu; x
\right)^2, \qquad x = \frac{1-\sqrt{1-4/c^2}}{2}, $$
but the issue is that this is only correct for $c>2$, and the rhs is
complex for $c<2$. Appell function would only be defined by analytic
continuation in this case anyway, and I didn't find anything useful about
non-principal branches of Appell or hypergeometric functions.
For $c>2$, Mathematica also gives the following:
$$
I(\nu,c) = \frac{\Gamma(\frac{1+3\nu}{2})c^{-1-2\nu}}{\Gamma(\frac{1-\nu}{2})\Gamma(1+\nu)^2}
\,\,\,{}_3F_2\left( \frac{1+\nu}{2}, \frac{1}{2}+\nu,
\frac{1+3\nu}{2}; 1+\nu, 1+2\nu; \frac{4}{c^2} \right), $$
but this is incorrect when $c<2$.
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
You may find a simple solution in (Almost) Impossible Integrals, Sums, and Series, Sect. 6.60, page $533$, showing that $$\sum_{i=1}^\infty \sum_{j=1}^\infty \frac{\Gamma(i) \Gamma(j) \Gamma(x)}{\Gamma(i+j+x)}=\frac{1}{2}\left(\psi^{(1)}\left(\frac{x}{2}\right)-\psi^{(1)}\left(\frac{x+1}{2}\right)\right),$$ where $\psi^{(1)}(x)$ is Trigamma function.