How can I prove this closed form for $\sum_{n=1}^\infty\frac{(4n)!}{\Gamma\left(\frac23+n\right)\,\Gamma\left(\frac43+n\right)\,n!^2\,(-256)^n}$

According to Mathematica, the sum is $$ \frac{3}{\Gamma(\frac13)\Gamma(\frac23)}\left( -1 + {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; -1\right) \right). $$

This form is actually quite straightforward if you write out $(4n)!$ as $$ 4^{4n}n!(1/4)_n (1/2)_n (3/4)_n $$ using rising powers ("Pochhammer symbols") and then use the definition of a hypergeometric function.

The hypergeometric function there can be handled with equation 25 here: http://mathworld.wolfram.com/HypergeometricFunction.html: $$ {}_3F_2\left(\frac14,\frac12,\frac34; \frac23,\frac43; y\right)=\frac{1}{1-x^k},$$ where $k=3$, $0\leq x\leq (1+k)^{-1/k}$ and $$ y = \left(\frac{x(1-x^k)}{f_k}\right)^k, \qquad f_k = \frac{k}{(1+k)^{(1+1/k)}}. $$

Now setting $y=-1$, we get the polynomial equation in $x$ $$ \frac{256}{27} x^3 \left(1-x^3\right)^3 = -1,$$ which has two real roots, neither of them in the necessary interval $[0,(1+k)^{-1/k}=4^{-1/3}]$, since one is $-0.43\ldots$ and the other $1.124\ldots$. However, one of those roots, $x_1=-0.436250\ldots$ just happens to give the (numerically at least) right answer, so never mind that.

Also, note that $$ \Gamma(1/3)\Gamma(2/3) = \frac{2\pi}{\sqrt{3}}. $$

The polynomial equation above is in terms of $x^3$, so we can simplify that too a little, so the answer is that the sum equals $$ \frac{3^{3/2}}{2\pi} \left(-1+(1-z_1)^{-1}\right), $$ where $z_1$ is a root of the polynomial equation $$ 256z(1-z)^3+27=0, \qquad z_1=-0.0830249175076244\ldots $$ (The other real root is $\approx 1.42$.)

How did you find the conjectured closed form?

This is a very interesting question. Since it is different from similar question of this kind in that a new technique is involved I will provide the answer. We have: \begin{eqnarray} &&\sum\limits_{n=1}^\infty \frac{(4n)!}{\Gamma(n+\frac{2}{3}) \Gamma(n+\frac{4}{3})(n!)^2 (-256)^n}=\\ && \frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!} \sum\limits_{n=1}^\infty \binom{4 n}{n} \cdot \underbrace{\frac{1}{n+\frac{1}{3}}}_{\int\limits_0^1\theta^{n-\frac{2}{3}} d\theta}\cdot \left( -\frac{3^3}{4^4}\right)^n=\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\int\limits_0^1 \theta^{-\frac{2}{3}} \cdot \left\{ \frac{x[\theta](1+\theta \frac{27}{256}x[\theta]^3)}{1+4\theta \frac{27}{256}x[\theta]^3}-1\right\} d\theta=\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{-\frac{4 \cdot 2^{\frac{2}{3}}}{3}\cdot \int\limits_1^{x[1]} \frac{1}{x^{\frac{4}{3}}} \cdot \frac{1}{(1-x)^{\frac{2}{3}}}dx-3\right\}\\ &&\frac{1}{(-\frac{1}{3})!}\frac{1}{(-\frac{2}{3})!}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\ &&\frac{\sqrt{3}}{2 \pi}\left\{4 \cdot 2^{\frac{2}{3}}(\frac{1-x[1]}{x[1]})^{\frac{1}{3}}-3\right\}\\ \end{eqnarray} In the first line we used elementary properties of factorials. In the second line we used my answer to Closed form solutions for a family of hypergeometric sums. . Here the function$x[\theta]$ is defined as a solution to the following polynomial equation: \begin{equation} 1-x[\theta] - \theta \frac{27}{256} x[\theta]^4=0 \end{equation} where out of the four different solutions we take the solution that is the closest to unity. In the third line we substituted for $x=x[\theta]$, in the fourth line we evaluated the integral and finally in the fifthe line we simplified the result. Here : \begin{equation} x[1]=\frac{2 \sqrt{2+\left(1+\sqrt{2}\right)^{2/3} \left(4 \sqrt{\frac{2}{1+\sqrt{2}-\sqrt[3]{1+\sqrt{2}}}}-2\right)}-2 \sqrt{2 \left(\left(1+\sqrt{2}\right)^{2/3}-1\right)}}{3 \sqrt[6]{1+\sqrt{2}}} \end{equation}