The expression mentioned in your question is a reflection of the fact that there are actually two "nice" extensions of the double factorial (one agreeing with it on even numbers and the other agreeing on odd numbers) and you can only combine them in a somewhat ad-hoc way.
Namely, the even and odd double factorials are
$$z!!_0 = \frac{2^{\frac{z}{2}} \Gamma\left(1+\frac{z}{2}\right)}{\Gamma\left(1+\frac{0}{2}\right)} = 2^{\frac{z}{2}} \Gamma\left(1+\frac{z}{2}\right)$$
and
$$z!!_1 = 1 \frac{2^{\frac{z-1}{2}} \Gamma\left(1+\frac{z}{2}\right)}{\Gamma\left(1+\frac{1}{2}\right)} = \sqrt{\frac{2}{\pi}} 2^{\frac{z}{2}} \Gamma\left(1+\frac{z}{2}\right).$$
The expression in your question is equal to
$$z!! = T_2(z) \: 2^{\frac{z}{2}} \Gamma\left(1+\frac{z}{2}\right),$$
where $T_2(z) = (\sqrt{2/\pi})^{(1-\cos (\pi z))/2}$ is equal to $1$ when $z$ is an even integer and to $\sqrt{2/\pi}$ when $z$ is an odd integer. I say that this is an ad-hoc construction because there are clearly many more possible $T_2(z)$ with this property (for example $(\sqrt{2/\pi})^{(1-\cos (13\pi z))/2}$).
Similarly, for the triple factorial we have three "nice" extensions, based on the triple factorials of numbers congruent to $0, 1, 2 \pmod 3$:
$$z!!!_0 = \frac{3^{\frac{z}{3}} \Gamma\left(1+\frac{z}{3}\right)}{\Gamma\left(1+\frac{0}{3}\right)},$$
$$z!!!_1 = \frac{3^{\frac{z-1}{3}} \Gamma\left(1+\frac{z}{3}\right)}{\Gamma\left(1+\frac{1}{3}\right)},$$
$$z!!!_2 = 2 \frac{3^{\frac{z-2}{3}} \Gamma\left(1+\frac{z}{3}\right)}{\Gamma\left(1+\frac{2}{3}\right)}.$$
So an extension of the triple factorial to $\mathbb{C}$ would be
$$z!!! = T_3(z) \: 3^{\frac{z}{3}} \Gamma\left(1+\frac{z}{3}\right),$$
where $T_3(z)$ is any function interpolating the three constants $1$, $3^{\frac{-1}{3}}/\Gamma\left(1+\frac{1}{3}\right)$ and $2 \cdot 3^{\frac{-2}{3}}/\Gamma\left(1+\frac{2}{3}\right)$ for $\mathbb{Z}\ni z=0,1,2 \pmod 3$ respectively, for example
$$T_3(z) = \left(\frac{3^{\frac{-1}{3}}}{\Gamma\left(1+\frac{1}{3}\right)} \right) ^{(1+2\cos(2\pi (z-1)/3))/3} \left(2 \frac{3^{\frac{-2}{3}}}{\Gamma\left(1+\frac{2}{3}\right)} \right)^{(1+2\cos(2\pi (z-2)/3))/3}.$$
You can see the resulting function here.
We can argue similarly for the higher-order multifactorials $z!^k$. The final expression is
$$z!^k = T_k(z) \: k^{\frac{z}{k}} \Gamma\left(1+\frac{z}{k}\right),$$
and a possible $T_k(z)$ generalizing the expressions for $T_2(z)$ and $T_3(z)$ given above is
$$T_k(z) = \prod_{j=1}^{k} \left(j\frac{k^{-j/k}}{\Gamma\left(1+\frac{j}{k}\right)} \right)^{E(k,j;z)},$$
where the exponent $E(k,j;z)=\frac{1}{k}\sum_{l=1}^k \cos\left(2\pi l \frac{(z-j)}{k}\right)$ is $1$ when $z = j \pmod k$ and vanishes when $z$ is any other integer. You can see its graph here (the slider also seems to allow fractional values of $k$, but I wouldn't trust that case).
Best Answer
Divide $x$ by $2$, then divide by $3$, and so on, until you cannot divide further. If you arrive at the number $1$, you started with a factorial. Otherwise you didn't.