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.
Best Answer
I don't know much about Mathematica and its working so can't comment on its failure to find a closed form for the Ramanujan type series in your question.
It is worth noting that the series is not one of those listed in Ramanujan's 1914 paper Modular equations and approximations to $\pi$. However Ramanujan's technique can be applied to evaluate the sum of series in closed form.
To that end let $k\in(0,1)$ be the elliptic modulus, $k'=\sqrt{1-k^2}$ be the complementary modulus and let $$K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{1}$$ be the complete elliptic integral of first kind. When $k$ is available from context we usually write $K, K'$ for $K(k), K(k') $ respectively and the variable $q=\exp(-\pi K'/K) $ is called the nome corresponding to $k$.
A lot of functions of the variable $q$ occurring naturally in elliptic function theory can be expressed in terms of the elliptic moduli and integrals. One such relevant function is Dedekind's eta function given by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{2}$$ Another function needed for our purpose here is the logarithmic derivative of eta function $$P(q)=24q\frac{d}{dq}\log\eta(q)=1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{3}$$ We have the closed form evaluations \begin{align} \eta(q)&=2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{4a}\\ \eta(q^2)&=\eta(\exp(-2\pi K'/K)) =2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{4b} \end{align} Differentiating the above formulas logarithmically with respect to $q$ and making use of $(3)$ as well as the formula $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{5}$$ we get $$2P(q^2)-P(q)=\left(\frac{2K}{\pi}\right)^2(1+k^2)$$ Replacing $k$ with $(1-k')/(1+k')$ changes $q$ into $q^2$ and $K$ into $(1+k')K/2$ (Landen transformation) and we get $$2P(q^4)-P(q^2)=\left(\frac{2K}{\pi}\right)^2\frac{1+k'^2}{2}$$ Putting $k=k'=2^{-1/2}$ so that $q=e^{-\pi} $ and elliptic integral $K=K(2^{-1/2})=\Gamma^2(1/4)/(4\sqrt{\pi})$ we get $$P(e^{-4\pi})=\frac{3}{2\pi}+\frac{3\Gamma^4(1/4)}{32\pi^3}\tag{6}$$ (using well known value $P(e^{-2\pi})=3/\pi$).
Next we need a series for $(2K/\pi)^2$ namely $$\left(\frac{2K}{\pi}\right)^{2} = \{1-(kk') ^2\} ^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1, 1; \frac{27g^{24}}{(4g^{24} + 1)^{3}}\right)\tag{7}$$ where $g=(2k/k'^2)^{-1/12}$. Using $k=3-2\sqrt{2}$ (singular modulus corresponding to nome $q=e^{-2\pi}$) we get the series mentioned in question $$\sum_{n=0}^{\infty}\frac{(6n)!}{n!^3(3n)!}\left(\frac{1}{287496} \right)^n=\frac{\sqrt{33}\Gamma(1/4)^4}{32\pi^3}$$ The calculation is greatly simplified if we observe that $$g^{24}=8,1-(kk')^2=1-k^2+k^4=k^2(4g^{24}+1)$$ and noting that $$K(k)=\frac{1+2^{-1/2}}{2}K(2^{-1/2})$$ via Landen transformation.
We can write using $(4b)$ $$\eta^4(q^2)=\frac{2^{-4/3}(kk')^{2/3}}{\sqrt{1-k^2+k^4}}\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}\left(\frac{27g^{24}}{(4g^{24}+1)^3}\right)^n$$ Next task is to differentiate the above series logarithmically with respect to $k$ and this is bit of a computational challenge. In the process we also make use of formula $(5)$. After a lot of tedious algebraic manipulation (readers should attempt verification only if they have a lot of spare time) one can obtain $$P(q^2)=\frac{(1-2k^2)(2-k^2)(1+k^2)}{2(1-k^2+k^4)^{3/2}} \sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(1+6n)\left(\frac{27g^{24}}{(4g^{24}+1)^3}\right)^n$$ Putting $k=3-2\sqrt{2}$ we get $$P(e^{-4\pi})=\frac{63}{11\sqrt {33}}\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(1+6n)\left(\frac{2}{11}\right)^{3n}$$ We had already computed the value of $P(e^{-4\pi})$ as $$P(e^{-4\pi})=\frac{3}{2\pi}+\frac{3}{\sqrt{33}}\sum_{0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}\left(\frac{2}{11}\right)^{3n}$$ and on comparing these values we get $$\frac{11\sqrt{33}}{4\pi}=\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(63n+5)\left(\frac{2}{11}\right)^{3n}$$ A major (and more difficult) part of Ramanujan's technique deals with evaluation of $P(q^2) $ in symbolic form as $$P(q^2)=\frac{3}{\pi\sqrt{n} } +\left(\frac{2K}{\pi}\right)^2f_n(k)$$ for $q=e^{-\pi\sqrt{n}}$ where $f_n(k) $ is some complicated algebraic function of $k$ dependent on $n$. Ramanujan gave explicit formulas for many values of $n$ including the case $n=4$ relevant here. We have instead taken a simpler route to evaluate $P(e^{-4\pi})$ using the well known value of $P(e^{-2\pi})$.
I also checked the paper linked in question and I find the claims made by its author regarding Ramanujan's technique being non-rigorous as simply a misconception. The development by Ramanujan in the theory of elliptic and theta functions is fully rigorous and is nothing more than processes of calculus applied in unconventional contexts and involves significant amount of computational labor. But it is a lot easier to understand his ideas and theories compared to the modern approach given by modular forms.
In particular let us observe that Ramanujan did not give details of computation, but the same goes for Borwein brothers and Chudnovsky brothers who didn't give details of their software based calculations. Berndt and his collaborators have been more forthcoming in their papers and give reasonable details of the calculations involved so that others can independently verify it if they have access to the needed software. Ramanujan's calculations were probably done on slate and never stored anywhere but their lack does not imply a lack of rigor.