Algebraic elements in $\mathbb{Q}_p$ are well studied and there was a question on mathoverflow about information regarding this.
I want to know following information:
Let $K_p=\overline{\mathbb{Q}}\cap \mathbb{Q}_p$. So we have extensions
$$\mathbb{Q} \subset K_p \subset \mathbb{Q}_p.$$ The first one is algebraic extension (of $\mathbb{Q}$) and next one is transcendental extension (of $K_p$).
Q. What is known about degrees of these extensions: $[K_p:\mathbb{Q}]$ and $[\mathbb{Q}_p:K_p]$?
To summarise some comments into an answer:
A. $[K_p:\Bbb Q]$ is countably infinite.
Namely, since the polynomial ring $\Bbb Q[x]$ is countable, so is $\overline{\Bbb Q}$ and so is $K_p$, so $[K_p:\Bbb Q]$ is at most countably infinite.
To see that it's not finite, notice the following: For $x \in \Bbb Z_p^\times$, $\Bbb Q_p$ contains an $n$-th root of $x$ for all $n$ which are coprime to $p(p-1)$. (E.g. by Hensel's lemma; it is maybe noteworthy that this characterises the set $\Bbb Z_p^\times$ within $\Bbb Q_p$, which can be used to prove e.g. that all automorphisms of $\Bbb Q_p$ are continuous hence trivial.)
Now let's for example look at the element $\alpha:=1+p \in \Bbb Z_p^\times \cap \Bbb Z$. Notice that $\alpha$ can be a perfect $\ell$-th power $k^\ell$ with $\ell$ prime and $k \in \Bbb Q$ (necessarily $k \in \Bbb N$), for at most finitely many primes $\ell_1, ..., \ell_r$, and hence for all $n$ with $gcd(n, 2\ell_1 ... \ell_r) =1$ the polynomial $x^n - \alpha$ is irreducible over $\Bbb Q$. So, for all $n$ with $gcd(n, 2\ell_1 ... \ell_r p(p-1)) =1$, the subextension $\Bbb Q(\sqrt[n]{\alpha}) \vert \Bbb Q$ of $K_p\vert \Bbb Q$ has degree $n$, where $\sqrt[n]{\alpha}$ denotes the $n$-th root of $\alpha$ existing in $\Bbb Q_p$ according to the previous paragraph. Since such $n$ can be arbitrarily high, $[K_p:\Bbb Q]$ must indeed be (countable) infinite.
B. $[\Bbb Q_p: K_p]$ is uncountable, more precisely, has the cardinality of $\Bbb Q_p$ which is $2^{\aleph_0}$.
This is just a cardinality argument using again that $K_p$ is countable, and so would be any countable-dimensional extension of it.