The way I read this, there can only be one round - a tie by the second player results in a loss as does every roll that is lower. The only alternative is when player 2 outright wins - binary event. Unless I'm misinterpreting?

I think you are reading it wrong. My manual solution for n = 3 matches Brad Warren's formula (19/27), and it has 6 possible scenarios:

Player 1 rolls a 3. Player 2 cannot beat a 3, so player 1 wins.

Player 1 rolls a 2. Player 2 can roll a 1 or 2 and lose, or roll a 3 to win, since player 1 cannot beat a 3.

Player 1 rolls a 1. Player 2 loses if he rolls a 1, and wins if he rolls a 3. If player 2 rolls a 2, player 1 rolls again. Player 1 will lose on 1 or 2, and will win on 3.

The same reasoning can obviously be expanded for any n, but I haven't modeled it with any generality. Theoretically there can be n "rounds" to a game, which would only occur if player 1 rolls 1, followed by player 2 rolling 2, then player 1 rolling 3, and so on.