Solution:
Answer: 2015.318180... In this solution, all logarithms will be taken in base 2. It is clear that $\log (Q)=\sum_{k=1}^{2012}\binom{2012}{k} \log (k)$. By paring $k$ with $2012-k$, we get $\sum_{k=1}^{2011} 0.5 * \log (k(2012-k))\binom{2012}{k}+\log (2012)$, which is between $0.5 * \log (2012) \sum_{k=0}^{2012}\binom{2012}{k}$ and $\log (2012) \sum_{k=0}^{2012}\binom{2012}{k}$; i.e., the answer is between $\log (2012) 2^{2011}$ and $\log (2012) 2^{2012}$. Thus $\log (\log (Q))$ is between $2011+\log (\log (2012))$ and $2012+\log (\log (2012))$. Also $3<\log (\log (2012))<4$. So we get $2014<M<2016$.