$$T(n) = \frac{\log_2 n}{\log_2\log_2 n}$$ is a near-solution of the difference equation for large $n$.  I expect this is the correct rate of growth and probably one can prove that by bounding the error term.
 
In fact there seems to be an asymptotic solution that starts
$$T(n) = \frac{\log_2 n }{\log_2\log_2 n }\left(1 + \frac{1}{\log\log_2 n} + \frac{2}{(\log\log_2 n )^2} + \cdots \right),$$
where I left off the subscript on the log a few times on purpose.