(cw answer based on Wojowu's link)

The only algebraic extension of $\mathbf{Q}$ that embeds into $\mathbf{Q}_p$ for all $p$ (or even for a density 1 set of primes) is $\mathbf{Q}$ itself.

For if $P\in\mathbf{Q}[t]$ is an irreducible polynomial of degree $\ge 2$, the set of primes $p$ for which $P$ has no root in $\mathbf{Q}_p$ has positive density: see this answer, which also applies to an arbitrary number field (=finite extension of $\mathbf{Q}$), where primes now mean primes of the number field.

The above statement roughly states that "the only algebraic number that belong to the intersection of all $\mathbf{Q}_p$ are rationals", which seems to answer your question. As already mentioned in comments, this statement doesn't really make sense, since there is no natural way to identify algebraic elements in different completions. An artificial way for it to makes sense is to mod out $\mathbf{Q}_p$ by the equivalence relation "having the same minimal polynomial (monic or zero) over $\mathbf{Q}$".