The point of this answer, building on a comment by "so-called friend Don", is to point out that behavior has density $0$ among all quintic extensions.

My guides here are the introductory portions of The geometric sieve and the density of squarefree values of invariant polynomials and Mass Formulae for Extensions of Local Fields and Conjectures on the Density of Number Field Discriminants, both by Bhargava.

Let $k$ be a field. Recall that an etale $k$-algebra is a direct sum of finitely many finite separable extension fields of $k$. (We'll be in characteristic zero, so you can ignore the adjective separable.) We'll write $\mathrm{Aut}(K/k)$ for the automorphism group of $K$ preserving $k$; this is a finite group. Let $K$ be a degree $p$-adic extension of $\mathbb{Q}_p$ of degree $n$. As I understand the philosophy of these papers, the probability of that the $p$-adic completion of a degree $n$ number field will be isomorphic to $K$ is supposed to be
$$\frac{p-1}{p} \frac{1}{\mathrm{Disc}_p(K)} \frac{1}{\# \mathrm{Aut}(K/\mathbb{Q}_p)}.$$

Moreover, these probabilities for distinct primes are supposed to be independent.

For $K$ an etale $\mathbb{Q}_p$ algebra, we write say that $K$ has symbol $(f_1^{e_1}, f_2^{e_2}, \ldots, f_r^{e_r})$ if $K \cong \bigoplus_{i=1}^r K_i$ where $K_i/\mathbb{Q}_p$ is a field extension of ramification degree $e_i$ and residue field extension of degree $f_i$. A lemma in the Mass Formula paper (Prop 2.1) allows us to group together any place where we sum over the set of all etale $\mathbb{Q}_p$-algebras with a given symbol.

So, our desired Euler factor is:

**The unramified extensions** These are symbols where all the $e_i$ are $1$; namely $(1,1,1,1,1)$, $(2,1,1,1)$, $(3,1,1)$, $(4,1)$, $(5)$. We compute:
$$\frac{p-1}{p} \left( \frac{1}{120} + \frac{1}{12} + \frac{1}{8} + \frac{1}{6} + \frac{1}{4} + \frac{1}{5} \right) = \frac{p-1}{p}.$$
It is not a coincidence that the sum came out to $1$; see equation (2.4) in the Mass formula paper.

**The other symbols** It is also okay to have symbol $(1^2, 1,1,1)$, $(1^2, 2,1)$, $(1^2, 1^2, 1)$ or $(2^2, 1)$. Each of these symbols corresponds to several possible etale $\mathbb{Q}_p$-algebras (for example, for $p$ an odd prime, there are $2$ ramified
quadratic extensions of $\mathbb{Q}_p$, so $(1^2, 1,1,1)$ describes two options), but Prop 2.1 in the Mass formula paper means we don't have to think about this. I compute that the respective terms are

$$\frac{p-1}{p} \left( \frac{1}{p} \left( \frac{1}{6} + \frac{1}{2} \right) + \frac{1}{p^2} \left( \frac{1}{2} + \frac{1}{2} \right) \right)$$

Putting everything together, our Euler factor is
$$\frac{p-1}{p} \left( 1+\frac{2}{3p} + \frac{1}{p^2} \right) = 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3}.$$

The point is that
$$\prod_p \left( 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3} \right) =0.$$
Now, Bhargava's paper doesn't directly allow us to us the prudct in this manner because his best theorem, Theorem 1.3 in the sieve paper, only applies when, for all sufficiently large $p$, we include all the terms with discriminant $1$ or $p$. However, let $p_N$ be the probability that the ramification is as you wish for all $p < N$. Then Bhargava's Theorem 1.3 does show that $p_N$ exists and equals $\prod_{p<N} \left( 1-\frac{1}{3 p} + \frac{1}{3 p^2} - \frac{1}{p^3} \right)$. So this product is an upper bound for the probability that a quintic behaves as desired for all primes. Sending $N \to \infty$, we see that the proportion of quintics with the desired behavior is $0$.