Analytic continuation of Dixon's identity

Question

Many well-known combinatorial identities has an analytic version. For example, the following identities $$ 2^n = \sum_{k=0}^n \binom{n}{k} $$ $$ \binom{2n}{n} = \sum_{k=1}^n \binom{n}{k}^2 $$ can be generalized to $$ 2^z = \sum_{n=0}^{\infty} \binom{z}{n} $$ and $$ \frac{\Gamma(1+2z)}{\Gamma(1+z)^2} = \sum_{n=0}^{\infty} \binom{z}{n}^2 $$ where $z$ is in a small neighborhood of zero in $\mathbb C$ and $\binom{z}{n}=\frac{z(z-1)\dots (z-n+1)}{n!}$. For the Dixon's identity, $$ \sum_{k=0}^n (-1)^k \binom{2n}{k}^3 = (-1)^n \frac{(3n)!}{(n!)^3}, $$ it is less obvious to guess the analytic version of it. After some trial and error, I found $$ \sum_{n=0}^{\infty} (-1)^n \binom{z}{n}^3 = \cos \left(\frac{\pi z}{2}\right) \frac{\Gamma(1+3z/2)}{\Gamma(1+z/2)^3} $$ To prove this, my idea is to find a functional equation satisfied by both sides of the formula. However, before delving into it, I want to ask if this formula exists somewhere in the literature and in which context.