Solution:

Every side of such a triangle is a cube edge, a face diagonal of length $\sqrt{2},$ or a space diagonal of length $\sqrt{3}.$ Only three shapes occur. A triangle of two adjacent edges and a face diagonal is right with area $\frac{1}{2};$ there are $4$ per face, or $24.$ A triangle of three face diagonals is equilateral with area $\frac{\sqrt{3}}{2};$ each is determined by the three vertices adjacent to one of the $8$ cube vertices, so there are $8.$ A triangle of an edge, a face diagonal, and a space diagonal is right with legs $1$ and $\sqrt{2},$ so its area is $\frac{\sqrt{2}}{2};$ each of the $4$ space diagonals forms one with each of the $6$ vertices off that diagonal, so there are $24.$ (Indeed $24 + 8 + 24 = \binom{8}{3} = 56.$)

The total area is $$24 \cdot \frac{1}{2} + 8 \cdot \frac{\sqrt{3}}{2} + 24 \cdot \frac{\sqrt{2}}{2} = 12 + 4\sqrt{3} + 12\sqrt{2} = 12 + \sqrt{48} + \sqrt{288},$$ so $m + n + p = 12 + 48 + 288 = 348.$