The key rule is that you can't remove a block from the uppermost complete level. We can prove (see below) that this means the top 3 levels will always have at least 6 blocks between them. Underneath that, the remaining blocks (up to 48 of them) could theoretically be just a single block per level, so this gives you an upper bound of 3 + 48 = 51 levels.
(I'm counting the uppermost row of blocks as a level even if it only ever contained one or two blocks; if you don't want to count the top level until it's complete then the answer is 50.)
Proofs:
First, we prove that the 2nd level is always a complete level of 3 blocks, by induction on the number of turns taken. At the beginning of the game (0 turns), it is true. Then after any turn on which it is true, there is one of two cases:
- The top level is incomplete. In this case, the 2nd level can't be modified and the top level will remain the top, so it will still be true after one more turn.
- The top level is complete. In this case, after one more turn that top level will become the (complete) 2nd level.
Now, we prove that the top 3 levels will always have at least 6 blocks. Assume towards contradiction that at some point there are fewer. Then since the 2nd level must have 3 by the lemma above, the 1st and 3rd level must each have only 1 block. But then that 3rd level with one block must, one turn earlier, have been an incomplete 2nd level, contradiction.