Disclaimer: This answer is building on a deleted answer by Lukas Heger. Also, a first version contained a wrong argument in the case $p=2$ which is hopefully fixed now.
Call $K := \mathbb Q_p(p^{1/n})$. The question is for what $n$ the extension $K\vert \mathbb Q_p$ is Galois. [Actually, a more precise question would be whether $K$ is well-defined as subfield of an algebraic closure of $\mathbb Q_p$. Surely we can define a field $\tilde K :=\mathbb Q_p [X] / (X^n-p)$, but when we embed it into a given algebraic closure $\overline {\mathbb Q_p}$, we have $n$ different choices where to map the residue of $X$ to, and in general different choices generate different field extensions; such an extension is normal hence Galois if and only if all choices define the same field.]
OP already remarked that if $n$ divides $p-1$, then $K$ is Galois via Kummer theory and the well-known fact that $\mathbb Q_p$ contains the $(p-1)$-th roots of unity. In case $p=2$, obviously $n=2$ still works. We will show now that these are the only possibilities.
So assume $K$ is Galois (i.e. well-defined as a subfield of $\overline{\mathbb Q_p}$). Now, $K \vert \mathbb Q_p$ is totally ramified of degree $n$. If $\zeta_k$ denotes a primitive $k$-th root of unity in $\overline {\mathbb Q_p}$, then the other roots of $X^n-p$ in $\overline {\mathbb Q_p}$ are $\zeta_n p^{1/n}, \zeta_n^2 p^{1/n}, ..., \zeta_n^{n-1} p^{1/n}$; so by being Galois, $K$ contains $E := \mathbb Q_p(\zeta_n)$. But $E\vert \mathbb Q_p$ has inertia degree $ > 1$ unless $n = p^r m$ with $m \vert (p-1)$ and $r \ge 0$.
Further, if $r \ge 1$, $K$ contains $\zeta_p$. Since $[\mathbb Q_p(\zeta_p) : \mathbb Q_p] = p-1$, we must have $m=p-1$, in particular $K$ contains a $(p-1)$-th root of $p$. On the other hand, with $\zeta_p$ the field $K$ also contains (1, 2) a $(p-1)$-th root of $\color{red}{-}p$; so it must contain a $(p-1)$-th root of $-1$, i.e. a $(2p-2)$-th root of unity.
For odd $p$, such a $\zeta_{2p-2}$ generates an extension of inertia degree $\ge 2$, contradiction. So $r=0$ and we are done.
For $p=2$, the above is still true but tells us nothing besides $n=2^r$; as it should, since obviously $r=1$ is allowed. To exclude all $r \ge 2$, let us look at the case $r=2$ first, i.e. $K=\mathbb Q_2(\sqrt[4]{2})$. Then $Gal(K\vert \mathbb Q_2)$ is of order $4$. It cannot be cyclic because $K$ has the two subfields $\mathbb Q_2(\zeta_4)$ and $\mathbb Q_2(\sqrt 2)$ which are distinct quadratic extensions of $\mathbb Q_2$ (4,5). So the Galois group would need to be $\simeq C_2 \times C_2$ i.e. contain three distinct elements of order $2$. Because the Galois group would act transitively on the roots of $X^4-2$, each of those three involutions would split those four roots into two pairs and then swap the two roots in each of the pairs. One computes from this that e.g. the element $\sqrt[4]{2} \cdot \zeta_4 \sqrt[4]{2} = \zeta_4 \sqrt{2}$ would be fixed by each element of the Galois group, i.e. be contained in $\mathbb Q_2$, which it obviously is not. So we have excluded $n=4$ i.e. $r=2$. [Compare the counterexample at the end of this answer.]
For $r\ge 3$, we can reduce to $r=2$ as follows: The general discussion showed that $K$ contains the cyclotomic field $E=\mathbb Q_2(\zeta_{2^r})$ which here is of degree $2^{r-1}$; so $[K:E]=2$. Since $E\vert \mathbb Q_2$ is known to be Galois, we see that $Gal(K\vert \mathbb Q_2)$ has a normal subgroup of order $2$. Such a subgroup is automatically central; but $E\vert K$ is actually cyclic, so $Gal(K\vert \mathbb Q_2)$ is cyclic modulo its centre, hence is surely abelian. By Galois theory this implies that each subfield of $K$ is Galois as well, including the subfield $K':= \mathbb Q_2(\sqrt[4]{2})$ which was the case $r=2$ we had lead to a contradiction above. So $r \in \{0,1\}$ and we are done.
Since the answer can be stated as: $\mathbb Q_p(p^{1/n}) \vert \mathbb Q_p$ is Galois if and only if $\mathbb Q_p$ contains the $n$-th roots of unity, I originally conjectured a general Galois-theoretic argument for
If $F$ is a field, $char(F) \nmid n$, and $a \in F$ such that $f(X):=X^n-a \in F[X]$ is irreducible, then $F[X]/(f)$ is a normal extension of $F$ if and only if $F$ contains the $n$-th roots of unity;
where "if" is the easy direction of Kummer theory. However, a counterexample to "only if" (which can be compared to the argument in the case $p=2, r=2$ above) is given by $F=\mathbb Q(i\sqrt2), a=2, n=4$, i.e. $K= F(\sqrt[4]{2})$ which is Galois over $F$ even though the only roots of unity in $F$ are $\pm 1$. In fact, $Gal(K\vert F)$ is not cyclic, but $\simeq C_2 \times C_2$; its nontrivial elements are the involutions defined by $\sigma: \sqrt[4]{2} \leftrightarrow i\sqrt[4]{2}$, $\tau: \sqrt[4]{2} \leftrightarrow -i\sqrt[4]{2}$, and $\sigma \tau : \sqrt[4]{2} \leftrightarrow -\sqrt[4]{2}$. Note that the fixed field of $\sigma \tau$ is $F(i)=F(\sqrt 2)$ which does contain the fourth root of unity $i=\zeta_4$, however the fixed fields of $\sigma$ and $\tau$ give two more quadratic subextensions.
Another counterexample, this time with a cyclic Galois group, is given by leoli1 in a comment.
