I think the following identity is true. How could we prove it?
$${_4F_3}\left(\begin{array}c 1,1,1,1 \\\tfrac54,2,2\end{array}\middle|\,1\right) + 3\,{_4F_3}\left(\begin{array}c\tfrac12,\tfrac12,1,1\\\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\,1\right) = \frac{\pi^3}{8} + \frac{ 3\pi^2}{4}\ln 2 - \frac 72 \zeta (3).$$
Here ${_pF_q}$ is the generalized hypergeometric function and $\zeta(3)$ is Apéry's constant.
A numerical approximation of the identity:
$$4.7994017718717349316710651835631714860755433336848275119882157\dots$$
$\def\Li{{\mathrm{Li}}}$These can be evaluated in closed form with a CAS (I used Mathematica).
The main tool is the (Euler) integral representation $$ F\left({a,\cdots\atop b,\cdots}\middle|\, z\right) = \int_0^1\frac{t^{a-1}(1-t)^{b-a-1}}{B(a,b-a)} F\left({\cdots\atop\cdots}\middle|\,zt\right)\,dt. $$
Call the hypergeometric values $F_1$ and $F_2$. Then $$ 4F_1 = \int_0^1 \frac{\Li_2(t)\,dt}{t(1-t)^{3/4}} = \int_0^1 \frac{\Li_2(1-t)\,dt}{(1-t)t^{3/4}} = \int_0^1 \frac{dt}{(1-t)t^{3/4}}\int_0^t \frac{du}{u}(-1)\log(1-u). $$ Then substitute $t=s^4$, interchange the order of summation, integrate over $u^{1/4} < s < 1$, and substitute $u=v^4$, giving $$ 4F_1 = -\int_0^1 \frac{16v^3\,dv}{1-v^4} \log v\left(\log\frac{1+v}{1-v} + i \log\frac{1-i v}{1+i v}\right). $$ After expanding the logs, and expanding the rational the rational function $\frac{v^3}{1-v^4}$ in partial fractions, each individual integrand has an elementary antiderivative (which a good CAS should be able to find). I omit the explicit antiderivatives, because they are very long. The final value is $$ 4F_1 = \tfrac{7}{8} \pi ^3-48 \Im\Li_3(\tfrac12+\tfrac i2)+\tfrac{3}{2} \pi (\log 2)^2+3\pi ^2 \log2-14\zeta(3). $$
The second hypergeometric is similar. Choosing $a$ and $b$ appropriately, and using a CAS to simplify the hypergeometric term in the integrand, then substituting $t=s^2$, we have $$ F_2 = \int_0^1 \frac{ds}{2\sqrt{1-s^2}}\big(\Li_2(s) - \Li_2(-s)\big). $$
Now, the same expression with a "$+$" instead of "$-$" can be evaluated by a CAS easily: $$ \int_0^1 \frac{dt}{2\sqrt{1-t^2}}\big(\Li_2(t)+\Li_2(-t)\big) = \tfrac1{48}\pi^3 - \tfrac14\pi(\log2)^2. $$
So it's only necessary to find the integral $$ \int_0^1 \frac{\Li_2(t)\,dt}{2\sqrt{1-t^2}}. $$ Using, like before, the integral representation $$ \Li_2(t) = -\int_0^t \frac{du}{u}\log(1-u), $$ we find that the integral is equal to $$ -\int_0^1 \frac{dt}{2\sqrt{1-t^2}} \int_0^t \frac{du}{u}\log(1-u) = \int_0^1 \frac{du}{2u}\log(1-u)\big(\arcsin u - \tfrac12\pi\big), $$ where we've done the integral over $t$ first.
Expanding the arcsine in logarithms to get $$ -\frac\pi4 \frac{\log(1-u)}{u} - \frac{i}{2} \frac{\log(1-u) \log(iu+\sqrt{1-u^2})}{u}, $$ my CAS was able to find a very long elementary (with polylogarithms) antiderivative for this expression. Then taking limits $u=0,1$ and subtracting gives the expression $$ \int_0^1 \frac{\Li_2(t)\,dt}{2\sqrt{1-t^2}} = \tfrac5{48}\pi^3 + 2\Im \Li_3(1-i). $$
Using the identity $$ 0 = - \Li_3(z) + \Li_3(1/z) - \tfrac16\log^3(-z)-\tfrac16\pi^2\log(-z) $$ with $z=1-i$, $1/z = \frac{1+i}{2}$, and putting things together gives $$ F_2 = -\tfrac1{32}\pi^3 + 4\Im \Li_3(\tfrac12+\tfrac{i}2)-\tfrac18\pi(\log2)^2. $$
Finally we reach $$ F_1 + 3F_2 = \tfrac18\pi^3 + \tfrac34\pi^2\log2 - \tfrac72\zeta(3). $$