The answer is countably infinite.
Suppose you have an alphabet $\Sigma$ with $n = |\Sigma|$ symbols. $\Sigma^k$ denotes the set of (exactly) $k$-length strings, and $\Sigma^* = \cup_{k=0}^\infty \Sigma^k$ denotes the set of finite strings over $\Sigma$. $\Sigma^0$ denotes the set containing the empty string, ie, $\Sigma^0 = \{ \epsilon \}$.
Order the symbols in $\Sigma = \{ \sigma_1, ..., \sigma_n \}$, and define $\nu : \Sigma \to \{0,...,n-1\}$ by $\nu(\sigma_k) = k-1$. Then define $\phi: \Sigma^* \to \mathbb{N}\cup \{0\}$ by $\phi(\epsilon) = 0$, and $\phi(\alpha_1 \cdots \alpha_l) = \frac{n^l -1 } {n-1}+\sum_{k=1}^l \nu(\alpha_k) n^{k-1}$, for $l \geq 1$. A bit of work shows that $\phi$ is a bijection, hence $\Sigma^*$ is countably infinite.
(Note that the bijection is 'almost' the base-$n$ value of the string, but must allow for the fact that, for example, the strings $0$ and $000$ are different while their base-$n$ value is the same. In the above, with $\Sigma = \{0,1\}$, $\phi(0) = 1$, $\phi(000) = 7$.)
Unfortunately, this trick doesn't generalize to countable alphabets.