I think I found an answer, but I would appreciate if someone would check to see if I did this correctly before I mark this question as answered.
Under Series formulas for the Incomplete Beta Function, Wolfram gives that
$$B(z; a, b) \propto \frac{z^a}{a}\left(1+O(z)\right)$$
and under Approximation for the Beta Function Wikipedia gives that
$$B(a, b) \propto \Gamma(b)a^{-b}$$
when $a$ grows and $b$ stays fixed (credit to Brevan Ellefsen for finding this second approximation). Because of these two approximations we can write our limit as
$$\lim_{n\to\infty} nx^{n-1}\frac{C\frac{\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\frac{n+1}{2}}}{\Gamma\left(\frac{1}{2}\right)\left(\frac{n+1}{2}\right)^{-\frac{1}{2}}}$$
where $C$ is some constant not depending on $n$.
$$\lim_{n\to\infty} nx^{n-1}\frac{C\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}}{\sqrt{\pi}\sqrt{\frac{n+1}{2}}}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}x^{n-1}\left(1-\frac{x^2}{4}\right)^{\frac{n+1}{2}}$$
for some other constant $c$.
Let $x = 2\sin\theta$. The limit then becomes
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\right)^{n-1}\left(\cos\theta\right)^{n+1}$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(2\sin\theta\cos\theta\right)^{n-1}\cos^2\theta$$
$$c\lim_{n\to\infty} \frac{n}{\sqrt{n+1}}\left(\sin\left(2\theta\right)\right)^{n-1}\cos^2\theta$$
Note that, whenever $\left|\sin\left(2\theta\right)\right| < 1$, the $\left(\sin\left(2\theta\right)\right)^{n-1}$ term tends to $0$ more quickly than the $\frac{n}{\sqrt{n+1}}$ term tends to $\infty$ (since $\frac{n}{\sqrt{n+1}}$ is asymptotically a power of $n$ and $\left(\sin\left(2\theta\right)\right)^{n-1}$ is exponential), and the limit is just $0$; also note that when $\left|\sin\left(2\theta\right)\right| = 1$ the limit tends to $\infty$. Thus, this reduces to the problem of finding $\theta$ such that
$$\sin^2\left(2\theta\right) = 1$$
$$4\sin^2\theta\cos^2\theta = 1$$
$$x^2\left(1-\frac{x^2}{4}\right) = 1$$
$$4x^2-x^4 = 4$$
$$x^2 = 2$$
$$x = \pm\sqrt{2}$$
So, at these points the limit tends to $\infty$ whereas everywhere else it tends to $0$.
let's consider $I(\alpha)=\int_0^1\frac{(1-t)^{-\alpha}-(1-t)^{\alpha}}{t}dt$, where $\alpha\in(0;1)$.
We can apply the regularisation and consider $I(\alpha)=\lim_{\epsilon\to 0}I_\epsilon(\alpha)=\lim_{\epsilon\to 0}\int_0^1\frac{(1-t)^{-\alpha}-(1-t)^{\alpha}}{t^{1-\epsilon}}dt$.
Then
$$I_\epsilon(\alpha)=B(\epsilon;1-\alpha)-B(\epsilon;1+\alpha)=\Gamma(\epsilon)\Big(\frac{\Gamma(1-\alpha)}{\Gamma(1-\alpha+\epsilon)}-\frac{\Gamma(1+\alpha)}{\Gamma(1+\alpha+\epsilon)}\Big)$$
Given that at $\epsilon\to0\,\,\,\epsilon\,\Gamma(\epsilon)=\Gamma(1+\epsilon)=1-\gamma\epsilon+O(\epsilon^2)\,\,\Rightarrow\,\,\Gamma(\epsilon)=\frac{1}{\epsilon}-\gamma+O(\epsilon)$,
and $\Gamma(1\pm\alpha+\epsilon)=\Gamma(1\pm\alpha)+\Gamma'(1\pm\alpha)\epsilon+O(\epsilon^2)=\Gamma(1\pm\alpha)\Big(1+\psi(1\pm\alpha)\epsilon+O(\epsilon^2)\Big)$
(where $\psi(z)=\frac{\Gamma'(z)}{\Gamma(z)}$ - digamma function)
$$I_\epsilon(\alpha)=\Big(\frac{1}{\epsilon}+O(1)\Big)\Big(\frac{1}{1+\psi(1-\alpha)\epsilon+O(\epsilon^2)}-\frac{1}{1+\psi(1+\alpha)\epsilon+O(\epsilon^2)}\Big)$$
$$=\frac{1}{\epsilon}\big(\psi(1+\alpha)\epsilon-\psi(1-\alpha)\epsilon\big)+O(\epsilon)$$
$$I(\alpha)=\lim_{\epsilon\to 0}I_\epsilon(\alpha)=\psi(1+\alpha)-\psi(1-\alpha)$$
Given that $\psi(1+\alpha)=\psi(\alpha)+\frac{1}{\alpha}$ and $\psi(1-\alpha)-\psi(\alpha)=\pi\cot\pi\alpha$ (digamma-function)
$$I(\alpha)=\frac{1}{\alpha}-\pi\cot\pi\alpha$$
Best Answer
Inspired by this answer, we write \begin{align}S&=\sum_{k\ge1}\frac{\operatorname B(k,1/2)}{(2k+1)^2}=\sum_{k\ge1}\frac{4^k}{k(2k+1)^2{2k\choose k}}=\sum_{k\ge1}\frac{\int_0^{\pi/2}\sin^{2k+1}x\,dx}{k(2k+1)}.\end{align} Notice that \begin{align}\sum_{k\ge1}\frac{x^{2k+1}}{k(2k+1)}&=\int_0^x\sum_{k\ge1}\frac{t^{2k}}k\,dt=-\int_0^x\log(1-t^2)\,dt\\&=2x-x\log(1-x^2)-2\operatorname{arctanh}x\end{align} through $\log(1-t^2)=\log(1-t)+\log(1+t)$, from which \begin{align}S&=\int_0^{\pi/2}\left(2\sin x-2\sin x\log\cos x-2\operatorname{arctanh}\sin x\right)\,dx\\&=2-2\int_0^1\log u\,du-2\int_0^{\pi/2}\log(\tan x+\sec x)\,dx\\&=4-4G.\end{align}