Trying to calculate this and it seems incredibly hard. I'm going by the rules of oragami, i.e. you can identify any edge with another edge (or crease) and any point where two edges meet to any other point where two edges meet. I'm actually not really sure on those rules, I just want each fold to be exactly reproducible and it seems like that would do it.

Lets just say for simplicity sake that after every fold you have to unfold and use the creases as edges. While taking into account symmetries would be cool, i.e. two sequences of folds are equivilant iff they paper can be rotated or reflected to create the other one, it's prob too hard to start off that way.

Any ideas about how to go about solving this?
for n=1.... 4 possible ways
for n=2.... a bunch

I think with a lot of these combinatorial problems, the trick is to find some way of representing the folds, and then dealing with those rather than dealing with the pictures themselves.

Perhaps it might make sense to have some sort of "level k fold" where k is somehow the count of the depth of folds (how many folds it takes to create this fold), and then try counting from there. It will at least help you to wrap your mind around the possibilities.

For example...

There are 4 level 0 folds (two diagonal and two orthogonal).
There are 14 level 1 folds (4 for each diagonal fold, and 3 for each orthogonal fold)
Etc.

Then to find out how many ways you can have two folds, you take two level 0 folds, or you take one level 0 and one level 1 fold.

Hope this gives you a good start. (For all I know, it will crash and burn at level 2 folds!)

This just got me thinking... maybe it's enough to know that you can make a new crease from any two preexisting creases uniquely. The problem is that sometimes you double count.
(Anyways this makes a relatively easy upper bound)
So when do you "accidentally" align two edges together when you are trying to align two other edges together?