The full etale fundamental groups in question are, I believe, complicated infinite profinite groups. (They are however "small" in the technical sense that they have only finitely many open normal subgroups of any given finite index, as follows from Hermite's finiteness theorem in algebraic number theory.)

The abelianization of $\pi_1(\operatorname{Spec}(\mathbb{Z}[\frac{1}{p}])$ is the maximal abelian extension of $\mathbb{Q}$ which is ramified only at $p$ (and infinity). By Class Field Theory, this is the direct limit of the ray class fields of conductor $p^n (\infty)$, i.e., the field generated by all $p$-power roots of unity. The Galois group is thus the inverse limit of the groups $(\mathbb{Z}/p^n \mathbb{Z})^{\times}$. When $p$ is odd, this is isomorphic to $\mathbb{Z}_p \times \mu_{p-1}$ (where the second factor is cyclic of order $p-1$). So yes, this depends on $p$!

The full etale fundamental groups in question are, I believe, complicated infinite profinite groups. (They are however "small" in the technical sense that they have only finitely many open normal subgroups of any given finite index, as follows from Hermite's finiteness theorem in algebraic number theory.)

The abelianization of $\pi_1(\operatorname{Spec}(\mathbb{Z}[\frac{1}{p}])$ is the Galois group of the maximal abelian extension of $\mathbb{Q}$ which is ramified only at $p$ (and infinity). By Class Field Theory, this field is the direct limit of the ray class fields of conductor $p^n (\infty)$, i.e., the field generated by all $p$-power roots of unity. The Galois group is thus the inverse limit of the groups $(\mathbb{Z}/p^n \mathbb{Z})^{\times}$. When $p$ is odd, this is isomorphic to $\mathbb{Z}_p \times \mu_{p-1}$ (where the second factor is cyclic of order $p-1$). So yes, this depends on $p$!