Can we count in depth $\frac{\lg n}{\lg \lg n}$?
Consider a fanin 2 circuit of $C$ depth $O(\log n)$. Divide the layers of $C$ into $O(\log n/\log\log n)$ blocks each of $\log\log n$ consecutive layers. We now wish to replace each block by a depth 2 circuit. Namely, each gate in the last layer of a block depends on at most $2^{\log\log n} = \log n$ gates of the last layer in the block below. We can thus replace each gate in the last layer by a DNF of polynomial size with inputs being the gates in the last layer of the block below. Doing this for all gates in the last layers for all blocks and connecting these should yield the desired circuit.

Let me note this is essentially the best one can obtain: the switching lemma allows for lower bounds all the way to depth $\log n/ \log\log n$.