How to compute the asymptotic growth of $\binom{n}{\log n}$?

I'm interested with tight bounds for: $$f(n)={n\choose{\log{n}}}$$ It sounds like it's something simple, but I can't get a nice expression I can use.

Any ideas on how to do this?

You can make use of Stirling's approximation.

$$ \log n! = n\log n - n + \frac{1}{2} \log 2\pi n + O\left(\frac{1}{n}\right)$$

$$\log \binom{n}{\log n} = \log n! - \log ((n-\log n)!) - \log ((\log n)!) $$

$$ = n\log n -(n-\log n)\log (n-\log n) + O(\log n \log \log n)$$

$$ = n \log n -(n - \log n)\left(\log n + \log \left(1 - \frac{\log n}{n}\right)\right) + O(\log n \log \log n)$$

$$ = \log^2 n + (n-\log n)\left(\frac{\log n}{n} + O\left(\frac{\log^2 n}{n^2}\right)\right) + O(\log n \log \log n)$$

$$ = \log^2 n + O(\log n \log \log n) $$

Thus your binomial coefficient is

$$ n^{\log n +O(\log \log n)}$$

We can make it more accurate by computing the leading terms in $O(\log n \log \log n)$.