Solution:

Pass to the dual cube: the octahedron's faces correspond to a cube's vertices, and two faces share an edge exactly when the corresponding cube vertices are adjacent. Following the numbers $1, 2, \ldots, 8$ and back to $1$ traces a closed $8$-step circuit through all the cube's vertices, and the requirement is that every step is a diagonal (an edge of one of the two inscribed tetrahedra, or one of the $4$ long space diagonals). There are $16$ such diagonals.

Each vertex lies on exactly one long diagonal, so the circuit cannot take two long diagonals in a row, and switching between the two tetrahedra is possible only via a long diagonal. Hence the circuit uses either $4$ long diagonals alternating with tetrahedron edges, or $2$ long diagonals separated by $3$-edge paths in each tetrahedron. In the first case, choosing a pair of opposite edges in each tetrahedron ($3 \cdot 2$ ways) gives $6$ octagons, each traceable as $8 \cdot 2$ permutations: $96.$ In the second case, a $3$-edge path in one tetrahedron can be chosen in $4! = 24$ ways, and the return path through the other tetrahedron is then forced up to $2$ choices, giving $8 \cdot 24 \cdot 2 = 384$ permutations.

So $96 + 384 = 480$ of the $8! = 40320$ labelings work, and the probability is $\frac{480}{40320} = \frac{1}{84}.$ Thus $m + n = 1 + 84 = 85.$