Solution:

A tetrahedron with equal opposite edges embeds in a rectangular box with the six edges as face diagonals. If the box has dimensions $a \times b \times c,$ then $a^2 + b^2 = 41,$ $a^2 + c^2 = 80,$ and $b^2 + c^2 = 89.$ Adding gives $a^2 + b^2 + c^2 = 105,$ so $(a, b, c) = (4, 5, 8).$ The box minus four corner tetrahedra of volume $\frac{abc}{6}$ each leaves $$V = abc - 4 \cdot \frac{abc}{6} = \frac{abc}{3} = \frac{160}{3}.$$

All four faces are congruent triangles with sides $\sqrt{41},$ $\sqrt{80},$ $\sqrt{89}.$ By Heron's formula in the form $16F^2 = 2(a^2b^2 + b^2c^2 + c^2a^2) - (a^4 + b^4 + c^4)$ applied to the squared sides $41, 80, 89,$ we get $16F^2 = 28098 - 16002 = 12096,$ so $F = \sqrt{756} = 6\sqrt{21}.$

The point equidistant from all four faces is the insphere center, and decomposing the tetrahedron into four pyramids over the faces gives $V = \frac{1}{3} r \cdot 4F.$ Hence $$r = \frac{3V}{4F} = \frac{160}{24\sqrt{21}} = \frac{20}{3\sqrt{21}} = \frac{20\sqrt{21}}{63},$$ and $m + n + p = 20 + 21 + 63 = 104.$