Explicit description of $\Bbb Q_p \cap \bar{\Bbb Q}$ Note that we can embed $\Bbb Q_p$ into $\Bbb C$, as it is discussed here. But as far as I understand, this embedding sends the power
series to transcendental elements, so we can't certainly embed $\Bbb Q_p$ into $\bar{\Bbb Q}$ (I love typing barBbbQ). And when we complete $\Bbb Q$ with the $p$-adic norm, then we get, for example, $(p-1)$-th roots of unity (to see this, just note that $x^{p-1} - 1$ factors in $\Bbb Z / p$, so by Hensel's lemma it factors in $\Bbb Q_p$).

So the question is: do we get anything else? What?
Tautologically, you want to describe the number fields $K \subset \overline{\mathbf{Q}}$ which embed in $\mathbf{Q}_p$. It's certainly not going to be possible to describe such fields completely. For example,
consider the polynomial:

$$f(x) = x(x-1)^{n-1} + p g(x),$$

where $g(x)$ is an arbitrary polynomial. For a random choice of $g(x)$, the polynomial will be irreducible and (generically, by Hilbert irreducibility) have splitting field $S_n$. On the other hand, Hensel's Lemma will ensure that $K = \mathbf{Q}[x]/f(x)$ embeds into $\mathbf{Q}_p$. 

For a Galois extension $K/\mathbf{Q}$, the condition is equivalent to asking that $K$ is unramified at $p$, and that the Frobenius element (conjugacy class) $\mathrm{Frob}_p \in G = \mathrm{Gal}(K/\mathbf{Q})$ is the identity.

For example, if $K/\mathbf{Q}$ is abelian, then the ramification condition implies that the conductor $N$ is prime to $p$, that is, $K \subset \mathbf{Q}(\zeta_N)$. The largest subfield of $\mathbf{Q}(\zeta_N)$ which embeds into $\mathbf{Q}_p$ corresponds, by Galois theory, to the largest quotient on which $\mathrm{Frob}_p$ is trivial. Yet

$$\mathrm{Frob}_p = [p] \in (\mathbf{Z}/N \mathbf{Z})^{\times} = G
= \mathrm{Gal}(\mathbf{Q}(\zeta_N)/\mathbf{Q}).$$

The corresponding extension has Galois group $(\mathbf{Z}/N \mathbf{Z})^{\times}/[p].$

For example:


$\mathbf{Q}(\zeta_N) \hookrightarrow \mathbf{Q}_p$ if and only if $[p]$ is trivial in $G$, or equivalently, if $p \equiv 1 \mod N$.
There will be infinitely many abelian fields of any degree inside $\mathbf{Q}_p$ --- to generate extensions of degree $M$, for example, choose $N$ to be divisible by at least two distinct primes of the form $1 \mod M$. Then $G$ surjects onto $(\mathbf{Z}/M \mathbf{Z})^2$, and hence $G/[p]$ surjects onto $\mathbf{Z}/M \mathbf{Z}.$
A version of the inverse Galois problem predicts that there exists (arbitary many) fields $K$ with Galois group any finite group $G$ in which any (fixed) prime $p$ splits completely.
The set $\mathbb Q_p\cap \overline{\mathbb Q}$ is simply the set of $p$-adic numbers which satisfy a polynomial with coefficients in $\mathbb Q$. In fact you can see the two sets living inside the same one, namely $\overline{\mathbb Q_p}$.