Prove that $_4F_3\left(\frac13,\frac13,\frac23,\frac23;1,\frac43,\frac43;1\right)=\frac{\Gamma \left(\frac13\right)^6}{36 \pi ^2}$
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.
Wow, amazing! Solved 9 years later! Thanks all for digging this up, and then for solving it. Can this give a general form for
$$_4F_3(\frac1m,\frac1m,\frac2m,\frac2m;\frac{m+1}m,\frac{m+1}m,1;1)$$
I should probably give some motivations for this. In the following paper, I looked at the expected exit time of a planar Brownian motion starting at 0 from a regular $m$-gon centered at 0:
https://projecteuclid.org/euclid.ecp/1465262013
It is (up to a constant which depends on the size of the polygon)
$$_4F_3(\frac1m,\frac1m,\frac2m,\frac2m;\frac{m+1}m,\frac{m+1}m,1;1)\times \frac{m^2}{\beta(1/m,(m-2)/m)^2},$$
which doesn't exactly roll off the tongue. However, for an equilateral triangle there is a different method for calculating this, and it gives $1/6$. So we get an identity by equating the two, and that is the identity. Now, the question is, can we use this method to get a nicer expression for the $_4F_3$ for larger $m$? This would then be a nicer expression for the expected exit time of Brownian motion from the regular $m$-gon.
A purely analytic (i.e. not probabilistic) version of this all can be found here, because the expected exit time is basically the Hardy H^2 norm of the domain, up to a constant.
https://arxiv.org/abs/1205.2458