I've calculated rigorously that the expectations for $n=1,2,3,4$ are $0,1,3,6$ respectively. For example, when $n=4$:

- One turn brings the balls to a 211 state (meaning 2 balls of one color and 1 ball each of two more colors).
- From a 211 state, there is a $2/5$ probability of going to another 211 state, a $2/5$ probability of going to a 31 state, and a $1/5$ probability of going to a 22 state.
- In other words, there is a $2/5$ probability of staying in the 211 state and a $3/5$ probability of leaving it; the expected number of turns it takes to leave the 211 state is thus $1/(1-2/5) = 5/3$. When it does leave, there's a $2/3$ probability of being in a 31 state and a $1/3$ probability of being in a 22 state.
- By Gambler's Ruin, the expected number of turns to go from the 22 state to the 4 state is $2(4-2)=4$, while the expected number of turns to go from the 31 state to the 4 state is $3(4-3)=3$.
- Therefore, the total expected number of turns for the $n=4$ game is $1 + 5/3 + (\frac23\cdot 3 + \frac13\cdot 4) = 6$.

Moreover, I've run simulations for $n=5,6,7$. The data strongly suggests that the expectations are $10,15,21$ respectively.

I am thus persuaded to conjecture that the expected stopping time for $n$ balls in general is exactly $\binom n2 = n(n-1)/2$.