The (tight) lower bound is $n + \lceil \lg n \rceil - 2$ (where $\lg n$ means $\log_2 n$).
I'll prove tightness first: that this can be achieved (apparently the idea is due to Lewis Carroll!). First find the maximum using a "tennis tournament" structure: first compare the $n$ elements in pairs, then compare the $n/2$ "winners" in pairs, and so on. (Unpaired elements get a bye to the next round.) Since every element except the winner loses exactly once, this takes $n-1$ comparisons. But now note that the second largest element must be one which lost to the winner, as it couldn't have been defeated by any other element. So you need to find the maximum among all the (up to) $\lceil \lg n \rceil$ elements that were defeated by the winner, and finding this maximum can be done in $\lceil \lg n \rceil - 1$.
We can prove this is a lower bound as well. Let the number of elements that lost to the maximum be $m$.
- Firstly, you need to find the maximum, since one cannot be sure some element is the second maximum without knowing which element is the maximum. Further, for each element that lost to the maximum ($m$ of them), this comparison was useless in determining whether it was or not the second maximum. In other words, all elements other than the maximum must lose at least once, and all but one ($m-1$) of the elements that directly lost to the maximum must lose at least once more, no matter in which order the comparisons were made. So we need at least $n-1 + m-1 = n + m - 2$ comparisons.
- To state the same thing differently: $n-2$ of the elements must be found less than the second-largest element — comparisons with the largest element do not help here — plus $m$ of them must lose directly to the maximum by definition, so we need at least $n - 2 + m = n + m - 2$ comparisons.
This proves the $n + \lceil \lg n\rceil - 2$ lower bound if we show that $m \ge \lceil \lg n \rceil$, i.e., an adversary can make sure you always have at least $\lceil \lg n\rceil$ elements that lost to the winner. This is proved here or here; an argument goes as follows: the adversary keeps for each element a list of elements known to be less than or equal to it. Whenever a query's result is already "known", the adversary gives that answer, else it declares the one which has a larger list the winner (breaking ties arbitrarily). The maximum is determined when the size of some list grows to n, and with each comparison the size of any list can only double, so the number of comparisons needs to be at least $\lceil \lg n\rceil$.