Exponential Time Classification
Does one add the number of operations or the complexity of each operation?


This is a good question! And the answer is, "it depends".

In any model of computation, some type of operation is considered "fundamental", and this operation is considered to run in $O(1)$. For example, some models consider multiplication to be fundamental. But others consider multiplication exponential in the number of bits.

No matter what, though, your algorithm is going to be doing $2^n$ decrement operations on the last element of the list, so it's always going to be $\Omega(2^n)$ (where the omega means "at least", as opposed to omicron's "at most").


  Is the complexity O(2**n) or is it O(1)*(2**(n+1)-2) reducing to constant time: O(1)?


I'm not sure I understand this part of the question. $O(1) \times (2^{n+1}-2)$ is $O(2^n)$, not $O(1)$.

However, your time complexity certainly isn't double-exponential. As you noted, you're doing $2^{n+1}-2$ decrement operations, so if you consider decrement to be constant, the total time is $O(2^{n+1}-2) = O(2^n)$.

Even if you did an additional $O(2^n)$ operation for each decrement, the time complexity would only be $O(2^n) \times O(2^n) = O(2^{2n})$.