Answer: 63

Solution:

Since $E$ is the midpoint of $\overline{AB},$ $AE = EB = 12.$ Let $P$ be the centroid, which trisects the medians: $AP = \frac{2}{3} \cdot 18 = 12$ and $PE = \frac{1}{3} \cdot 27 = 9.$ By the power of the point $E$ with respect to the circumcircle, $EF \cdot EC = EA \cdot EB = 144,$ so $EF = \frac{144}{27} = \frac{16}{3}.$

Triangle $AEP$ is isosceles with $AE = AP = 12$ and base $PE = 9,$ so the altitude from $A$ to $\overline{PE}$ is $\sqrt{144 - \frac{81}{4}} = \frac{3\sqrt{55}}{2},$ giving $[AEP] = \frac{1}{2} \cdot 9 \cdot \frac{3\sqrt{55}}{2} = \frac{27\sqrt{55}}{4}.$ Since $F$ and $P$ both lie on line $CE,$ triangles $AEF$ and $AEP$ share the apex $A$ and have collinear bases, so $$\begin{aligned} [AEF] &= \frac{EF}{EP}\,[AEP] \\ &= \frac{16/3}{9} \cdot \frac{27\sqrt{55}}{4} \\ &= 4\sqrt{55}. \end{aligned}$$

Finally, since $E$ is the midpoint of $\overline{AB},$ $[AFB] = 2\,[AFE] = 8\sqrt{55},$ and $m + n = 8 + 55 = 63.$