How many $p$-adic numbers are there?

The field $\mathbb Q_p$ is the fraction field of $\mathbb Z_p$.

Since you already know that $|\mathbb Z_p|=2^{\aleph_0}$, let us show that this is also the cardinality of $\mathbb Q_p$:

Note that every element of $\mathbb Q_p$ is an equivalence class of pairs in $\mathbb Z_p$, much like the rationals are with respect to the integers.

Since $\mathbb Z_p\times\mathbb Z_p$ is also of cardinality continuum, we have that $\mathbb Q_p$ can be injected into this set either by the axiom of choice, or directly by choosing representatives which are co-prime.

This shows that $\mathbb Q_p$ has at most continuum many elements, since $\mathbb Z_p$ is a subset of its fraction field, then the $p$-adic field has exactly $2^{\aleph_0}$ many elements.

To complete Student's answer : $\mathbb Z_p \subset \mathbb Q_p \subset \mathbb C_p$, and since $\mathbb C_p$, the algebraic closure of $\mathbb Q_p$, is isomorphic to $\mathbb C$, the field of the complex numbers, we know that $$ |\mathbb R| = |\mathbb Z_p| \le |\mathbb Q_p| \le |\mathbb C_p| = |\mathbb C| = |\mathbb R| \quad \Longrightarrow \quad |\mathbb Q_p| = |\mathbb R|. $$ by the Cantor-Bernstein theorem, which states that if $|A| \le |B|$ and $|B| \le |A|$ then $|A| = |B|$.

EDIT : One REALLY more easy way to see this (in the sense that I can assume less non-trivial things) is that we get an injection from $\mathbb R$ to $\mathbb Q_p$ (and vice-versa) by choosing one representation for each element on each side and associate them in the following manner : $$ \sum_{k=-\infty}^n a_k p^k = \underset{\in \mathbb R}{x} \longleftrightarrow \underset{\in \mathbb Q_p}{y} = \sum_{k=-n}^{\infty} a_{-k} p^k $$ (i.e. you flip the number over) which doesn't give us a bijection but rather two surjections. (If I wanted to have a bijection with this I would have trouble with the possible two representations of a number on each side.) Thus I can apply Cantor-Bernstein.