A detailed proof.
Let $S=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^3}$ be the sum we want to evaluate.
If we consider the binomial expansions
\begin{equation}
2-2\sqrt{1-x}=\sum_{n=0}^{\infty}\binom{2n}{n}{x^{n+1}\over 4^{n}(n+1)}\tag{1},
\end{equation}
\begin{equation}
-4\log{2}+4\log{\left(1+\sqrt{1-x}\right)}-4\sqrt{1-x}+4=\sum_{n=0}^{\infty}{\binom{2n}{n}x^{n+1}\over 4^n(n+1)^2}\tag{2},
\end{equation}
where $(1)$ can be obtained from the binomial expansion of $(1-x)^{-1/2}$ after multiplying both sides by ${1 \over x}$, then integrating in $(0,x)$, and $(2)$ doing the same as with $(1)$.
The change $x=k^{2}\sin^{2}x$ in $(2)$ gives
\begin{equation}
4\log{\left({\sqrt{1-k^{2}\sin^{2}x}\over 2}+{1\over 2} \right)}-4\sqrt{1-k^{2}\sin^{2}x}+4=\sum_{n=0}^{\infty}{\binom{2n}{n}k^{2n+2}\sin^{2n+2}(x)\over 4^{n}(n+1)^{2}}\tag{3},
\end{equation}
The next step we do is integrate $(3)$ in $(0,{\pi\over 2})$ and apply Wallis formula to obtain
\begin{align}
2\pi\log\left({\sqrt{1-k^{2}}+1\over 4}\right)+4\int_{0}^{\pi \over 2}{x\cos{x}\over \sin{x}\sqrt{1-k^{2}\sin^{2}{x} }}dx-4E(k)+2\pi \\ \nonumber ={\pi\over 2}\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}k^{2n+2}(n+1/2)\over 16^{n}(n+1)^{3}}\tag{4},
\end{align}
where $E(k)$ is the complete elliptic integral of the second kind.
Now we observe that
$$S=2S_{1}-S_{2},$$
where
$$S_{1}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 16^{n}(n+1)^{2}}\hspace{.5cm} and\hspace{.5cm} S_{2}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}(2n+1)\over 16^{n}(n+1)^{3}}.$$
First we will see that $S_{1}={16\over \pi}-4$. See Calculate $\sum_{n = 0}^\infty \frac{C_n^2}{16^n}$ .
Finally setting $k=1$ in $(4)$
$$S_{2}=\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}(2n+1)\over 16^{n}(n+1)^{3}}={4\over \pi}\left(-4\pi\log{2}+4\int_{0}^{\pi\over 2}{x\over \sin{x}}dx+2\pi-4\right) \\=-16\log{2}+{32G\over \pi}-{16\over \pi}+8,$$
where we have used the well known integral $\int_{0}^{\pi\over 2}{x\over \sin{x}}dx=2G$, being $G$ Catalan's constant. Then the desired sum is
\begin{align*}
S=2S_{1}-S_{2}=16\log{2}-{32G\over \pi}+{48\over \pi}-16.
\end{align*}
Bonus
\begin{equation}_4F_3\left(a;b;x\right)=\sum_{n=0}^{\infty}{(a_{1})_{n}(a_{2})_{n}(a_{3})_{n}(a_{4})_{n}x^{n}\over (b_{1})_{n}(b_{2})_{n}(b_{3})_{n}n!}
\end{equation}
where $(a)_{n}$ is the Pochammer's symbol. If $x=1$:
\begin{equation}
\sum_{n=0}^{\infty}{\left({1\over 2}\right)^{2}_{n}(1)_{n}^{2}\over(2)_{n}^2(3)_{n} n!}=2\sum_{n=0}^{\infty}{\binom{2n}{n}^{2}\over 2^{4n}(n+2)(n+1)^{3}}=32\log{2}-{64G\over \pi}+{608\over 9\pi}-24,
\end{equation}
\begin{equation}
\sum_{n=0}^{\infty}{\left({3\over 2}\right)^{2}_{n}(1)_{n}^{2}\over(2)_{n}(3)_{n}^2 n!}=16\sum_{n=1}^{\infty}{\binom{2n}{n}^{2}\over 2^{4n}n(n+1)^{2}}=64\log{2}-{128G\over \pi}-{320\over \pi}+96,
\end{equation}
where $G$ is Catalan's constant.
\begin{equation}
\sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over ({1\over 2})_{n}(1)_{n}({3\over 2})_{n}n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(2n+1)}={2\over \pi}+{\sqrt{2\pi}\over 2\Gamma{(3/4)}^2}-{\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}}.
\end{equation}
\begin{equation}
\sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over ({1\over 2})_{n}^2(2)_{n}n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(n+1)}={20\over 9\pi}+{\sqrt{2\pi}\over 9\Gamma{(3/4)}^2}+{2\sqrt{2}\Gamma{(3/4)}^2\over 3\pi^{3/2}},
\end{equation}
\begin{equation}
\sum_{n=0}^{\infty}{({1\over 4})_{n}^{2}({3\over 4})_{n}^{2}\over(1)_{n} ({3\over 2})_{n}^2n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(2n+1)^{2}}={8\over \pi}-{4\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}},
\end{equation}
\begin{equation}
\sum_{n=0}^{\infty}{(-{1\over 4})_{n}^{2}({1\over 4})_{n}^{2}\over({1\over 2})_{n} (1)_{n}^2n!}=\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(4n-1)^{2}}={2\over \pi}+{\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}},
\end{equation}
\begin{align}
\sum_{n=0}^{\infty}{(-{3\over 4})_{n}^{2}(-{1\over 4})_{n}^{2}\over(-{1\over 2})_{n}^{2} (1)_{n}n!}={9\over 64}\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}(n+1)^{2}}+1={11\over 6\pi}+{\sqrt{2} \Gamma{(1/4)}^2\over12\pi^{3/2}}+{3\sqrt{2}\Gamma{(3/4)}^2\over 2\pi^{3/2}}-{3\sqrt{2} \Gamma{(1/4)}^2 \Gamma{(3/4)}^4\over 8 \pi^{7/2}}.
\end{align}
\begin{equation}
\sum_{n=0}^{\infty}{\binom{4n}{2n}^{2}\over 256^{n}}\left({4\over (4n-1)^2}-{1\over (2n+1)^2}\right)={8\sqrt{2}\Gamma{(3/4)}^2\over \pi^{3/2}},
\end{equation}
