How does Hofmann elimination work in this reaction?
Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure.

For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring.



Once you do this, you see that there are only 2 hydrogen atoms that are anti-periplanar for an E2 elimination. One is on the axial methyl group (red). The other is on the N-ethyl group (blue). Only these two can easily be deprotonated for elimination. Of these two hydrogen atoms, the one on the axial methyl is very sterically crowded. Specifically, any base has to get around the ring structure and the encroaching flag pole hydrogen atoms. On the other hand, the hydrogen atom on the ethyl group sees no such steric encumberance.

Of the three elimination products, I expect to see:


Almost exclusively ethylene
Maybe a small amount of the methylidenecyclohexane
No methylcyclohexene


EDIT:
@Martin-マーチン was kind enough to compute the structure below. Notice the structural features we've already discussed: the ring conformation and the position of the anti-periplanar hydrogen atoms.