As described by Guillera in "Ramanujan Series with a Shift", one nice thing about Ramanujan-type $1/\pi^m$ formulas is by "shifting" them, they can yield a second value which may also have a closed-form. For $m>1$, only 12 non-trivial ones are known and we will apply the technique to the first half. Let $(x)_n$ be the Pochhammer symbol.
$$F_1(k) = \small\sum_{n=0}^\infty \frac{\big(\tfrac12+k\big)_n^5}{\big(1+k\big)_n^5}\,\left(20(n + k)^2 + 8(n + k) + 1\right)\,\left(\frac{-1}{\;2^2}\right)^n$$
then $\,F_1(0) = \dfrac{8}{\pi^2}\,$ and $\,F_1\big(\tfrac12\big) = 7\,\color{red}{\zeta(3)}.$
$$F_2(k) = \small\sum_{n=0}^\infty \frac{\big(\tfrac12+k\big)_n^5}{\big(1+k\big)_n^5}\,\left(820(n + k)^2 + 180(n + k) + 13 \right)\,\left(\frac{-1}{\,2^{10}}\right)^n$$
then $\,F_2(0) = \dfrac{128}{\pi^2}\,$ and $\,F_2\big(\tfrac12\big) = 256\,\color{red}{\zeta(3)}.$
$$F_3(k) = \small\sum_{n=0}^\infty \frac{\big(\tfrac12+k\big)_n^3\,\big(\tfrac13+k\big)_n\,\big(\tfrac23+k\big)_n}{\big(1+k\big)_n^5}\,\left( 74(n + k)^2 + 27(n + k) + 3 \right)\,\left(\frac34\right)^{3n}$$
then $\,F_3(0) = \dfrac{48}{\pi^2}\,$ and $\,F_3\big(\tfrac12\big) = \dfrac{16\pi^2}{3}.$
$$F_4(k) = \small\sum_{n=0}^\infty \frac{\big(\tfrac12+k\big)_n^3\,\big(\tfrac14+k\big)_n\,\big(\tfrac34+k\big)_n}{\big(1+k\big)_n^5}\,\left( 120(n + k)^2 + 34(n + k) + 3 \right)\,\left(\frac1{2^4}\right)^n$$
then $\,F_4(0) = \dfrac{32}{\pi^2}\,$ and $\,F_4\big(\tfrac12\big) = \dfrac{16\pi^2}{3}.$ (Same as above.)
$$F_5(k) = \small\sum_{n=0}^\infty \frac{\big(\tfrac12+k\big)_n^7}{\big(1+k\big)_n^7}\,\left(168(n + k)^3 + 76(n + k)^2 + 14(n + k) + 1 \right)\,\left(\frac1{2^6}\right)^{n}$$
then $\,F_5(0) = \dfrac{2^5}{\color{blue}{\pi^3}}\,$ and $\,F_5\big(\tfrac12\big) = \dfrac{\pi^4}{2}.$
The first relation was found by Boris Gourevitch around 2002. See this post.
$$F_6(k) = \small\sum_{n=0}^\infty\, s(n)\,\left( 43680(n + k)^4 + 20632(n + k)^3 + 4340(n + k)^2 + 466(n + k) + 21 \right)\,\left(\frac1{2^{12}}\right)^{n}$$
then $\,F_6(0) = \dfrac{2^{11}}{\color{blue}{\pi^4}}\,$ and $\,F_6\big(\tfrac12\big) = \dfrac{2^{10}\pi^4}{15},\,$ where $s(n) = \dfrac{\big(\tfrac12+k\big)_n^7\,\big(\tfrac14+k\big)_n\,\big(\tfrac34+k\big)_n}{\big(1+k\big)_n^9}.$
The first relation was found by Jim Cullen around 2010. See this post. After 2010, it seems no higher Ramanujan-type $1/\pi^m$ formula has been discovered.
Question: In Guillera's cited paper on $1/\pi$, some of the second values are conjectured. Likewise, some of the first and second values in this post are also conjectured. Anyone can prove a pair, especially if it involves $\zeta(3)$?