Your integral has an elementary closed form, that was correctly stated by Cleo in her answer without proof:
$$S=\int_0^1\left({_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)\right)^2dx=\frac{8\sqrt2+4\ln\left(\sqrt2-1\right)}{3\pi}.\tag1$$
Proof: Using DLMF 14.3.6 we can express the hypergeometric function in the integrand as the Legendre function of the $1^{st}$ kind (also known as the Ferrers function of the $1^{st}$ kind) with fractional index:
$${_2F_1}\!\left(-\tfrac14,\tfrac54;\,1;\,\tfrac{x}2\right)=P_{\small1/4}(1-x).\tag2$$
Now the integral can be written as
$$S=\int_0^1\left(P_{\small1/4}(1-x)\right)^2dx=\int_0^1\left(P_{\small1/4}(x)\right)^2dx.\tag3$$
To evaluate it, we use formula 7.113 on page 769 in Gradshteyn & Ryzhyk:
$$\int_0^1P_\nu(x)\,P_\sigma(x)\,dx=\\\frac{\frac{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac{\pi\nu}2\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(1+\frac\nu2\right)}{\Gamma\left(\frac12+\frac\nu2\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac{\pi\nu}2\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\nu)(\sigma+\nu+1)}.\tag4$$
Note that in our case $\nu=\sigma=\frac14$, so we cannot use this formula directly because of the term $(\sigma-\nu)$ in the denominator. Instead, we let $\nu=\frac14$ and find the limit for $\sigma\to\frac14$:
$$S=\lim\limits_{\sigma\to{\small1/4}}\int_0^1P_{\small1/4}(x)\,P_\sigma(x)\,dx=\\\lim\limits_{\sigma\to{\small1/4}}\frac{\frac{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}\sin\!\left(\frac{\pi\sigma}2\right)\cos\!\left(\frac\pi8\right)-\frac{\Gamma\left(\frac12+\frac\sigma2\right)\,\Gamma\left(\frac98\right)}{\Gamma\left(\frac58\right)\,\Gamma\left(1+\frac\sigma2\right)}\sin\!\left(\frac\pi8\right)\cos\!\left(\frac{\pi\sigma}2\right)}{\frac\pi2(\sigma-\frac14)(\sigma+\frac54)}.\tag5$$
To evaluate the limit, we use l'Hôpital's rule. This gives quite a big expression that I will not copy here. It contains values of the gamma and digamma functions at rational points, that could be simplified to elementary terms using the Gauss digamma theorem and identities given in the MathWorld and in the famous Vidūnas paper, yielding the desired result $(1)$.
Indeed, we can have a more general result:
$$\int_0^1\left({_2F_1}\!\left(-\nu,\nu+1;\,1;\,\tfrac x2\right)\right)^2dx=\int_0^1\left(P_\nu\left(x\right)\right)^2dx=\\\frac{1+\!\left[\psi_0\!\left(1+\frac\nu2\right)-\psi_0\!\left(\frac12+\frac\nu2\right)\right]\frac{\sin(\pi\nu)}\pi}{1+2\nu}.\tag6$$
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
By DLMF 15.6.1 $$ 2F\left({1/2,1/2\atop3/2};1\right)=\int_0^1\frac{\mathrm dt}{\sqrt{t(1-t)}}=\operatorname{B}(1/2,1/2)=\Gamma^2(1/2)=\pi. $$