$n+1$ rounds are the smallest number to guarantee a win. To see that $n+1$ rounds are sufficient, place $n$ of the men in a circle, each facing one woman in a concentric circle. The $n+1$st plays no part yet. Pair the couples who are facing one another. It any of them are matched, remove them from the circle. Then rotate one of the circles one position counterclockwise, and repeat. After $n-1$ rounds, each woman left in the circle has been pair with $n-1$ different men, each of who is not her match, so there is only one man left in the circle who can be her match, and she will face him in the $n$th round. Any woman not matched in round $n$ must be matched with the $n+1$st man, so $n+1$ rounds suffice.
I now believe that the necessity part of this proof is incorrect. I'm going to leave in my attempt at it, in hopes that someone can fix it.
To see that $n+1$ rounds are necessary, imagine that the producers of the show take an adversary approach. That is, they don't decide on the matches in advance, but make them up as the game goes along, in order to prolong the game as long as possible. Of course, at the end of the game, they must be prepared to announce all the matches, and they cannot contradict prior statements: once they say a couple isn't matched, it must stay unmatched.
In my original post, I described an adversarial strategy rather loosely, and the OP has rightly objected that I haven't demonstrated that no contradiction can arise. Here is a more sophisticated procedure, and a proof of correctness. The producers will use a flow network whose vertices are the men and women. At the start, there will be an edge of capacity $1$ from the source to each woman, an edge of capacity $1$ from each woman to each man, and an edge of capacity $1$ from each man to the sink. Clearly, the maximum flow in the graph is $n$. It is well-know that since all the capacities are integral, there is a maximum flow with integral flows, in this case, $0$ or $1$. As usual, we consider a man and woman as matched if the edge joining them has a flow of $1$. As the game progresses, the producers will modify the network to prevent any contradictions from arising.
I will prove that in each round the maximum flow in the network is equal to the number of unmatched women remaining, so that a complete matching is always possible. We will deal with the question of the "exceptional woman" later.
At each round, the producers examine the edges joining the current couple in turn. (If the couple has been paired previously, there is no such edge, as we shall see). For each such edge $e$, the producers run the max-flow algorithm to test whether the removal of edge $e$ from the graph will reduce the maximum flow. If not, $e$ is erased. If so, then the removal of $e$ can only reduce the flow by $1$, and a maximum flow must include a flow in $e$. The man and woman joined by $e$ are matched and removed from the graph. We now have a graph with one less woman, and a maximum flow of $1$ less, so the invariant is preserved.
Now the problem is to show that the players can't force this procedure to terminate prior to round $n+1$. I haven't been able to prove this, or anything close to it. Furthermore, I can show by example that it isn't true for $n=2$.
Suppose $n=2$ and let the women be A and B, and the men be a,b, and c. In the first round the players form the couples Aa, Bb. If the producers match both of these, the game is over, and if they match one, then the remaining woman pairs with c in round $2$ and the game ends. So, the producers make no match. In the second round, the players form the couples Ab, Ba. Now if the producers match both, the game is over, and if they match neither, then both a and b are unmatched, so the producers match Ab, say, then a is unmatched. Therefore, the producers cannot extend the game to three rounds. I think this may just be an anomaly and that my claim is true for $n>2$.
I find it hard to believe that this procedure doesn't last for at least $n$ rounds in general, but I haven't been able to prove it. I need to show that after $k$ rounds, there are at least $n-k$ unmatched women, or something like it, but I haven't been able to.