Had me worried for a moment! I thought the OP's question was
"how many possible auctions are there (starting from scratch)" - and I was thinking, something preposterous like
Graham's Number 
- that is, until I read the OP more carefully!
Starting from 7
♥, doesn't look overly difficult. Let's consider the cases where 7
♥ - with all its doubling options - is the final contract. There are seven cases: passed, two ways of being doubled (by E or W), then four ways of being doubled and redoubled.
Now for each of these seven cases, there are three ways in which the next bid of 7
♠ can occur. Thereafter there are, once again, seven pass/double/redouble combos. That makes (7 x 3 x 7) possible ways in which 7
♠ becomes the contract.
Also there are an equal number of cases where the next bid is 7NT (bypassing 7
♠). Another (7 x 3 x 7)
Finally, for each 7
♠ case, there are another 3 x 7 cases where the final outcome is 7NT. (7 x 3 x 7 x 3 x 7) in all.
So, to tot up:
7 +
(7 x 3 x 7) +
(7 x 3 x 7) +
(7 x 3 x 7 x 3 x 7)
On my calculator this comes to......
3388 ..... bingo!