It is not hard to prove that every element of $\mathbb{F}_p$ has a square root in $\mathbb{F}_{p^2}$: take any $a \in \mathbb{F}_p$ and consider the polynomial $f = X^2 - a$. If $f$ has a root in $\mathbb{F}_p$, then we are done. Otherwise $f$ is irreducible over $\mathbb{F}_p$, let $\beta \in \overline{\mathbb{F}_p}$ be a root of $f$, then the extension $\mathbb{F}_p(\beta)$ has degree $2$ over $\mathbb{F}_p$, and hence by the uniqueness of finite fields we have $\mathbb{F}_p(\beta) = \mathbb{F}_{p^2}$, so $\beta \in \mathbb{F}_{p^2}$. With exactly the same proof we can see that every element of $\mathbb{F}_p$ has a cube root in $\mathbb{F}_{p^3}$.
But I don't know if a similar statement holds for an arbitrary $n$. If $n>3$, the fact that $X^n - a$ has no root in $\mathbb{F}_p$ is no longer equivalent to being irreducible. So the thing that we would need to show in this case is that $X^n - a$ has an irreducible factor whose degree is a divisor of $n$. I don't see how to do that.
The problem would be solved if we could prove that $\mathbb{F}_{p^n}$ contains an element $\beta$ whose multiplicative order is $n(p-1)$, since in that case $\beta^n$ is a primitive root mod $p$. But, from the fact that $\mathbb{F}_{p^n}^{\times}$ is a cyclic group, if follows that the previous condition is equivalent to $n \mid p^{n-1} + p^{n-2} + \ldots + p + 1$, which does not necessarily hold (it does in some special cases, such as $p \equiv 1 \pmod{n}$).
Could someone please help with this?
I think that the following extension to prof. Lubin's argument settles the question in the affirmative.
First let's write $n=n_1n_2$, where all the prime factors of $n_1$ are also factors of $p-1$, and $\gcd(n_2,p-1)=1$. For all $a\in \mathbb{F}_p$ the equation $x^{n_2}-a$ has a root $y$ in the prime field, so it suffices to show that $y$ has an $n_1$th root in $\mathbb{F}_{p^{n_1}}\subseteq\mathbb{F}_{p^n} $.
Lemma. Assume that $q$ is a prime, and that the finite field $K$ contains a primitive $q^{th}$ root of unity $\zeta$. Let $\alpha\in K$ be arbitrary. Then the polynomial $$f(x)=x^q-\alpha$$ has a root in the unique degree $q$ extension $L$ of $K$.
Proof. If $f(x)$ has a root in $K$, then that root is also in $L$. If no such root exists in $K$, then such a root $\beta$ exists in $\overline{K}$. Because the other zeros of $f(x)$ are gotten from $\beta$ by multiplying it with a power of $\zeta$, we see that $K[\beta]$ is the splitting field of $f(x)$. Let $\sigma$ be a generator of the cyclic Galois group $\operatorname{Gal}(K[\beta],K)$. Then $\sigma(\beta)=\beta\zeta^\ell$ for some exponent $\ell$ coprime to $q$. As $q$ is a prime and $\zeta$ is fixed by $\sigma$, we see that $\sigma$ is of order that is a multiple of $q$. Therefore its order is exactly $q$, and we can conclude that $K[\beta]=L$. Q.E.D.
The claim follows easily from the Lemma. If $q$ is any prime factor of $n_1$, then $q\mid p-1$, so the prime field already contains the necessary roots of unity. Hence so do all its extensions, and the Lemma bites. More precisely, if $d\mid dq\mid n_1$, where $q\mid p-1$ is a prime, then, by induction hypothesis the equation $$ x^d=a $$ has a solution $x\in\mathbb{F}_{p^d}$. By the Lemma, the equation $y^q=x$ has a solution $y\in\mathbb{F}_{p^{dq}}$, and $y$ is then a solution of $$y^{dq}=a.$$ Repeating this step enough many times gets us to $n_1$ settling the claim.