I am fairly sure that asymptotically $$\frac{\sum \_{i=0}^{\lfloor n/2 \rfloor} {2(n-2i) \choose n-2i} {n \choose 2i} {4i \choose 2i}}{2^{3n - 1}} \sim \frac{2}{\pi n}.$$

I tried 

> 
> Limit[n\*Sum[Binomial[2n-4i,n-2i]\*Binomial[n,2i]\*Binomial[4i,2i],{i,0,n/2}]/2^(3n-1), n-> Infinity]
> 
> 
> 

but Mathematica can't compute it.

Is there another way to persuade Mathematica to do this sort of asymptotics?

In [Find asymptotics of $\sum\limits\_{i=0}^{n/3} 2^i \binom{n-i-1}{\frac{2n}{3}-1}$](https://mathematica.stackexchange.com/questions/32950/find-asymptotics-of-sum2ibinomialn-i-1-2n-3-1-i-0-n-3?rq=1) there is a clever use of Normal@Series as well as some other ideas for getting asymptotics but I didn't manage to get them to work for this problem.