Here's a 'straightforward' (i.e. non-rigorous) partial answer that may help you build your intuition. Assume all logs are base 2. By expanding $m$ times you can find that
$T(n) = 2^m T(n/2^m) + n\sum_{j=0}^{m-1} \log\log(n/2^j)$
so if we let $n=2^k$ and $m=k$, then
$T(n) = n T(1) + n\sum_{j=0}^{\log n - 1} \log\log (n/2^j)$
This gives you the easy bound
$\begin{align} T(n) & < n T(1) + n\sum_{j=0}^{\log n-1} \log \log n \\ & < nT(1) + n\log n \log\log n \end{align}$
and so you definitely have $T(n)=O(n\log n \log \log n)$.
By being more careful when bounding the sum (e.g. see Marvis's answer) you will be able to prove that $T(n)=\Theta(n\log n\log \log n)$ as well.