Given a cube, its cross-section (polygon) can have at most 6 edges. This can be seen from the fact that: The number of edges of a cross-section can't exceed the number of faces of the polyhedron.
However, this seems to be a fairly loose upper bound. Consider a regular Octahedron, its cross-section could also have at most 6 edges.
(1) Is there a better upper bound for this number? (2) Can this be generalized from polyhedron to polytope?
As already mentioned in a comment to that other refered to question, your quest is just about the Petrie polygon of the polyhedron. (Or, in fact, the rectification thereof, i.e. connecting its edge-centers each.)
The Petrie polygon (which happens to be a skew polygon) can be derived as follows: start at any vertex of your polyhedron and run along one of its incident edges. Continue on a neighbouring edge of one of its incident faces. Now swap over to the other neighbouring face of that second edge and continue along the neighbouring edge there. Continue doing so (swapping over and use the next edge there) until you get back to your starting vertex. - The Petrie polygon thus outlines nothing but the shadow rim of your chosen polyhedron (if it was a convex one and is to be oriented accordingly).
And indeed, this Petrie polygon also could be defined for higher dimensional polytopes too. Within 3D the rule said always go along 2 consecutive edges of a face, but no 3. Within 4D the adjusted rule would add to the former to go along 3 consecutive edges of a cell, but no 4. And for each further dimension add one more such restriction.
--- rk