$$T(n) = \begin{cases}T\left (n/\log_2{n}\right) + 1 & n>2\\1&n\leq 2\end{cases}$$
I'm tempted to think that the answer is $\Theta(\log n)$, but the denominator decreases as $n$ decreases. Some computer aided trials suggest that $T(n)$ grows slower than $\log_k n$ for any constant $k$