Despite this being a 10-year old post, it does come up when searching "double factorial analytic continuation", so I thought I'd expand on Ted's answer. (I'm new to StackExchange so hopefully I don't do anything wrong!)
The equation given by Ted (and Brad's linked Wikipedia page) does work for any complex input $x$, but it is specifically based on the double-factorial definition that starts with an odd integer: $$(2n - 1)!! = \prod_{k=1}^n (2k - 1) = 3 \cdot 5 \cdot 7 \cdots (2n - 1)$$ However, it seems quite common to extend the definition to even numbers like so: $$(2n)!! = \prod_{k=1}^n (2k) = 2 \cdot 4 \cdot 6 \cdots (2n)$$
It turns out that the analytic continuation for the 'even version' is different to the 'odd version'! I'll denote the odd version $F_-(x)$ and the even version $F_+(x)$.
The formula for $F_+(x)$ is actually quite simple to derive. We simply notice that the even double-factorial is a normal factorial except with each term doubled: $$(2n)!! = \prod_{k=1}^n 2k = 2(1) \cdot 2(2) \cdot 2(3) \cdots 2(n) = 2^n \cdot n!$$ Then to define $F_+$ we just set $2n = x$ and substitute in the gamma function: $$F_+(x) = 2^{x/2} \cdot \Gamma \left(\frac x2 + 1 \right)$$ Recalling the formula for $F_-(x)$: $$F_-(x) = \frac{2^{(x+1)/2}}{\sqrt \pi} \cdot \Gamma \left( \frac{x + 2}{2} \right)$$ We can then find the relationship between $F_+$ and $F_-$: $$F_+(x) = F_-(x) \cdot \sqrt{\pi / 2}$$ So one is a scaled version of the other!
It should be possible to define a more complicated function that interpolates between these two cases -- matching $F_+$ for even integers and $F_-$ for odd ones -- but at that point I expect it would be difficult to call any particular choice of interpolation 'the most intuitive'.
For 'higher-order' multifactorials, I expect they require more functions -- $3$ for $x!!!$, $4$ for $x!!!!$, etc. -- though I have not yet tried to confirm this.