The short answer is yes.
Firstly, you said when you have Bernoulli random variables that take values $-1, 1,$ then their sum converges to a normal distribution (presumably you mean by the Central Limit Theorem). This is slightly false; actually, the normalized sum will converge. In particular, if your variables are $X_1, X_2, \dots,$ then $$\frac{1}{\sqrt{n}} \sum_{i=1}^n X_i \stackrel{d}{\to} \mathcal{N}(0, \text{Var}(X)).$$
The Central Limit Theorem in general states that if you have a sequence of independent random variables $X_1, X_2, \dots$ that have the same distribution, so that $\sigma^2 := \text{Var}(X_i) < \infty$ and $\mu := \mathbb{E}[X_i]$ is the expected value, then $$\frac{1}{\sqrt{n}} \sum_{i=1}^n (X_i - \mu) \stackrel{d}{\to} \mathcal{N}(0, \sigma^2).$$
So indeed, this still works for the "3-valued Bernoulli variables" you mentioned. To know the variance of the normal distribution that the normalized sum converges to, you just need to compute the variables of your "3-valued Bernoulli variables."