The general procedure to solve for a MSNE in a 3-by-3 (or larger) game is always a bit tricky and involves some trial and error
- Step 1: Conjecture (i.e. guess) a subset of strategies that will be used in equilibrium
- Step 2: Calculate their probabilities using the indifference condition
- Step 3: Verify that the equilibrium payoff cannot be unilaterally improved upon; that is, no player has a strict incentive to deviate to another strategy
Suppose your conjectured strategies are $\{B,C\}\times\{A,B\}$ (it doesn't really matter what the basis for your conjecture is; you're going to find out one way or another whether that's correct). Next, calculate the probabilities using players' indifference conditions. Let $p=\sigma_1(B)$ and $q=\sigma_2(A)$, we have
\begin{align}
-3p&=-1&&\Rightarrow\quad p=1/3\\
3q+1-q&=1-q&&\Rightarrow\quad q=0.
\end{align}
[This suggests that your calculation for $q$ was incorrect.]
Lastly (this is the most easily forgotten step), check that no one has an incentive to deviate from this equilibrium. In this case, player 1's payoff is $1$, which is already highest given player 2's strategy of choosing $B$ with probability 1. He'd be indifferent between mixing in other proportions over $B$ and $C$, and his payoff is strictly lower if he plays $A$ with positive probability.
Player 2's expected payoff in this equilibrium is $-1$, which is also the highest given player 1's mixed strategy. She's indifferent between mixing over $A$ and $B$ with any other proportions and is strictly worse off if $C$ is played with positive probability.
So, one MSNE is $((0,1/3,2/3),(0,1,0))$. This is only borderline consistent with your initial conjecture because $\sigma_2(A)=0$. But it's nonetheless a MSNE. In fact, there are infinitely many MSNEs of this form: $((0,p,1-p),(0,1,0))$ where $p\ge1/3$. This is a full description of all equilibria (including the pure one) in this game.