Answer: 29

Solution:

Let the probability of face $F$ be $\frac{1}{6} + x,$ so the face opposite $F$ has probability $\frac{1}{6} - x$ (the six probabilities must sum to $1$). Since opposite faces sum to $7,$ a total of $7$ occurs exactly when the two dice show a pair of opposite faces. Of the six ordered pairs that sum to $7,$ four use only ordinary faces, and two pair $F$ with its opposite. Thus $$\begin{aligned} \frac{47}{288} &= 4\left(\frac{1}{6}\right)^2 \\ &\quad {}+ 2\left(\frac{1}{6} + x\right)\left(\frac{1}{6} - x\right) \\ &= \frac{1}{6} - 2x^2. \end{aligned}$$

Since $\frac{1}{6} = \frac{48}{288},$ this gives $2x^2 = \frac{1}{288},$ so $x = \frac{1}{24}.$ The probability of face $F$ is $\frac{1}{6} + \frac{1}{24} = \frac{5}{24},$ and $m + n = 5 + 24 = 29.$