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?
A complete answer now.
If we exploit the identities
$$\frac{4^n}{(2n+1)\binom{2n}{n}}=\int_{0}^{\pi/2}\sin(x)^{2n+1}\,dx \tag{1}$$
$$\frac{\arcsin(x)}{\sqrt{1-x^2}}=\frac{1}{2}\sum_{n\geq 1}\frac{4^n x^{2n-1}}{n\binom{2n}{n}},\qquad \arcsin^2(x)=\frac{1}{2}\sum_{n\geq 1}\frac{(4x^2)^n}{n^2\binom{2n}{n}}\tag{2}$$
we get:
$$(\pi-2)=\int_{0}^{\pi/2}\theta^2\sin(\theta)\,d\theta = \frac{1}{2}\sum_{n\geq 1}\frac{16^n}{(2n+1)n^2 \binom{2n}{n}^2}=\frac{1}{2}\sum_{n\geq 0}\frac{16^n}{(2n+3)(2n+1)^2\binom{2n}{n}^2} $$
and in a similar way:
$$\begin{eqnarray*}\frac{7\pi}{9}-\frac{40}{27}=\int_{0}^{\pi/2}\theta^2\sin^3(\theta)\,d\theta=\frac{1}{2}\sum_{n\geq 1}\frac{4^n 4^{n+1}}{n^2 (2n+3)\binom{2n}{n}\binom{2n+2}{n+1}}\end{eqnarray*}$$
If we integrate $\arcsin^2(x)$ and exploit $(1)$, we get:
$$ \sum_{n\geq 1}\frac{16^n}{(2n+1)^2 n^2 \binom{2n}{n}^2} = 4(\pi-3) $$
and maybe it is enough to integrate $\arcsin^2(x)$ once more to get a closed expression for the series of interest:
$$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}. $$
In such a case it appears a dependence on a dilogarithm, arising from the primitive of $\frac{\arcsin x}{x}\sqrt{1-x^2}$. At the moment I do not know if that is manageable or not, I have to carry out further experiments. Probably a logarithm appears from $\int_{0}^{\pi/2}\theta\cot(\theta)\,d\theta=\frac{\pi}{2}\log(2).$
Now that the path to the answer is a bit more clear, let us put $(1)$ and $(2)$ in a slightly more convenient way:
$$ \int_{0}^{\pi/2}\sin(x)^{2n+3}\,dx = \frac{4^{n}(2n+2)}{(2n+3)(2n+1)\binom{2n}{n}}\tag{1bis}$$
$$\arcsin^2(x)=\frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+2}}{(2n+2)(2n+1)\binom{2n}{n}}\tag{2bis}$$
If we integrate both sides of $(2\text{bis})$ we get:
$$ -2x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x) = \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)(2n+2)(2n+1)\binom{2n}{n}}\tag{3}$$
We just have to gain an extra $\frac{1}{(2n+3)}$ factor. For such a purpose, we divide both sides of $(3)$ by $x$ and perform termwise integration again, leading to:
$$ -4x+2\sqrt{1-x^2}\arcsin(x)+x\arcsin^2(x)+2\int_{0}^{\arcsin(x)}\frac{u\cos^2(u)}{\sin(u)}\,du\\= \frac{1}{2}\sum_{n\geq 0}\frac{4^{n+1} x^{2n+3}}{(2n+3)^2(2n+2)(2n+1)\binom{2n}{n}}\tag{4}$$
Now we evaluate both sides of $(4)$ at $x=\sin\theta$ and exploit $(1\text{bis})$ to perform $\int_{0}^{\pi/2}(\ldots)\, d\theta$.
That leads to:
$$ \sum_{n\geq 0}\frac{16^n}{(2n+3)^3(2n+1)^2\binom{2n}{n}^2}=(\pi-4)+\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u\cos^2(u)}{\sin(u)}\,du\,d\theta\tag{5} $$
and we may start buying beers, since the last integral boils down to $\int_{0}^{\pi/2}\int_{0}^{\theta}\frac{u}{\sin u}\,du\,d\theta$, that is well-known. We get:
$$\boxed{\begin{eqnarray*}\phantom{}_4F_3\left(1,1,1,\frac{3}{2};\frac{5}{2},\frac{5}{2},\frac{5}{2};1\right)&=&27\sum_{n\geq 0}\frac{16^n}{(2n+3)^3 (2n+1)^2 \binom{2n}{n}^2}\\&=&\color{red}{\frac{27}{2}\left(7\,\zeta(3)+(3-2K)\,\pi-12\right)}\end{eqnarray*}}\tag{6}$$
where $K$ is Catalan's constant. Please, do not ask me to do the same for other values of $\phantom{}_4 F_3$.
However, this instantly goes in my best of collection.
Addendum (15/08/2017) This result, together with another interesting identity relating $\phantom{}_4 F_3$ and $\text{Li}_2$, is going to appear on Bollettino UMI. You may have a glance at it on Arxiv.
Best Answer
Let $S$ be the given $_4F_3$, then (first equality comes from termwise integration), $$\begin{aligned} S &= -\frac{1}{9}\int_0^1 t^{-2/3} (\log t) {_2F_1}(2/3,2/3;1;t)dt =-\frac{1}{9} \frac{d}{da} \left(\int_0^1 t^{-2/3+a} {_2F_1}(2/3,2/3;1;t)dt \right)_{a=0}\\ &= -\frac{1}{9}\frac{d}{da}\left(\frac{\, _3F_2\left(\frac{2}{3},\frac{2}{3},a+\frac{1}{3};1,a+\frac{4}{3};1\right)}{ a+1/3}\right)_{a=0} \end{aligned}$$
It is easily seen $A=\sqrt{\pi } \Gamma \left(\frac{7}{6}\right)/\Gamma \left(\frac{5}{6}\right)^2$ is the value of the $_3F_2$ at $a=0$ (Dixon). Set $$\begin{aligned} &{d_{2/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3} + a,\frac{2}{3},\frac{1}{3};1,\frac{4}{3};1)} \right)_{a = 0}} \qquad {d_1} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3};1 + a,\frac{4}{3};1)} \right)_{a = 0}} \\ &{d_{1/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3} + a;1,\frac{4}{3};1)} \right)_{a = 0}} \qquad {d_{4/3}} = \frac{d}{{da}}{\left( {{_3F_2}(\frac{2}{3},\frac{2}{3},\frac{1}{3};1,\frac{4}{3} + a;1)} \right)_{a = 0}}\end{aligned}$$
By multivariable chain rule, $$S = A -\frac{1}{3}(d_{1/3}+d_{4/3})\tag{*}$$
In general, derivative of $_pF_q$ with respect to a parameter is intractable. One can only handle them in an ad hoc manner. In our situation, it is well-known that $_3F_2$ at $1$ satisfies certain transformations: two generators are the 1st and 3rd entry here. Using these two entries, we obtain $$\begin{aligned} & \quad _3F_2\left(\frac{2}{3},\frac{2}{3},a+\frac{1}{3};1,a+\frac{4}{3};1\right) \\ &= \frac{\Gamma \left(\frac{2}{3}\right) \Gamma \left(a+\frac{4}{3}\right) \, _3F_2\left(\frac{1}{3},\frac{2}{3},\frac{2}{3}-a;1,\frac{4}{3};1\right)}{\Gamma \left(\frac{4}{3}\right) \Gamma \left(a+\frac{2}{3}\right)} \\ &= \frac{\Gamma \left(\frac{2}{3}\right) \, _3F_2\left(a+\frac{1}{3},a+\frac{2}{3},a+\frac{2}{3};a+1,a+\frac{4}{3};1\right)}{\Gamma \left(\frac{2}{3}-a\right) \Gamma (a+1)} \\ &= \frac{\Gamma \left(-\frac{1}{3}\right) \Gamma \left(a+\frac{1}{3}\right) \Gamma \left(a+\frac{4}{3}\right) \, _3F_2\left(\frac{1}{3},\frac{2}{3},\frac{2}{3};\frac{4}{3},a+1;1\right)}{\Gamma \left(\frac{1}{3}\right)^2 \Gamma \left(a+\frac{2}{3}\right) \Gamma (a+1)}+\frac{\Gamma \left(\frac{1}{3}\right) \Gamma \left(a+\frac{1}{3}\right) \Gamma \left(a+\frac{4}{3}\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(a+\frac{2}{3}\right)^2} \end{aligned}$$
Observe that for all four $_3F_2$ above, their arguments are all like $(2/3,2/3,1/3;1,4/3)$, the only difference is $a$ appears at different places. This reveals why $(2/3,2/3,1/3;1,4/3)$ is special.
Introduce an operational definition: write $x\equiv y$ if $x-y$ is a "linear combination of gamma factors". For example, $x\equiv y$ if $x-y = A$. Now take derivative at $a=0$, we obtain $$\tag{**}d_{1/3}+d_{4/3} \equiv -d_{2/3} \equiv d_{1/3}+2d_{2/3}+d_1+d_{4/3} \equiv -d_1$$ Solving this system gives $$d_1 \equiv d_{2/3} \equiv d_{1/3}+d_{4/3} \equiv 0$$
Thus $d_{1/3}+d_{4/3}$ can be expressed into gamma function, so can $S$ according to $(*)$.
There is no difficulty in making $(**)$ explicit: $$d_{1/3}+d_{4/3}=\left(3-\frac{\pi }{\sqrt{3}}\right) A-d_{2/3}=d_1+d_{1/3}+2 d_{2/3}+d_{4/3}+\frac{1}{6} A \left(\sqrt{3} \pi -9 \log (3)\right)=-d_1+\frac{1}{2} A \left(\pi \sqrt{3}-6+3 \log (3)\right)+\frac{3 \left(3 \sqrt{3}-2 \pi \right) \Gamma \left(\frac{1}{3}\right)^2 \Gamma \left(\frac{7}{6}\right)^2}{\sqrt[3]{2} \pi ^2}$$
Solving gives $d_{1/3}+d_{4/3} = \dfrac{2 \sqrt{\pi } \left(27-4 \sqrt{3} \pi \right) \Gamma \left(\frac{13}{6}\right)}{21 \Gamma \left(\frac{5}{6}\right)^2}$. We also obtain values of $d_1, d_{2/3}$ as by-products.