Let $ p(a,b) $ denote the probability of obtaining $ a $ on the first die and $ b $ on the second. Then the probability of obtaining a sum of 7 is $$p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).$$Let the probability of obtaining face $ F $ be $ (1/6)+x $. Then the probability of obtaining the face opposite face $ F $ is $ (1/6)-x $. Therefore $$\begin{aligned}{{47}\over{288}}&= 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) \left({1\over6}-x\right)\cr&= {4\over36}+2\left({1\over36}-x^2\right)\cr&= {1\over6}-2x^2.\end{aligned}$$Then $ 2x^2=1/288 $, and so $ x=1/24 $. The probability of obtaining face $ F $ is therefore $ (1/6)+(1/24)=5/24 $, and $ m+n=\boxed{29} $.