This seems to hold:
This is just me of course writing out as elegantly as I know how the connection to OEIS A288780 I already proposed starting in the "negative space" discussion. I hope I did not make a mistake in translating my spreadsheet calculations to paper. Do I need to establish a clean proof the equality holds for all n, or is this the type of thing that would already be shown in some combinatorics book? I'm hoping anyway it will shed some understanding on minimal superpermutation lengths. Compare the left hand side to Greg Egan's upper bound of:
n! + (n-1)! + (n-2)! + (n-3)! + n - 3
So I propose we might see each individual permutation as contributing at least one to the minimal superpermutation length, giving you n! to start, and then the remaining length could be contributed by my left hand side, which if shown always equivalent to the right hand side would allow us to see the remaining length as being built up from differences of factorials. The right hand side has a "build from the ground up" perspective instead of skipping to (n-3)!
Explicitly, then, I am hopeful we can prove the minimal superpermutation length, or if we come up short there then maybe a new lower bound, to be:
n! + (n-1)! + (n-2)! + (n-3)! + 2