I believe that $\pi(\text{Spec } \mathbf Z[1/p])$ is the Galois group of the maximal extension of $\mathbf Q$ unramified outside $p$.

On the other hand, I believe that $\pi(\text{Spec }\mathbf Z[X])$ is trivial. If $Y$ were finite étale over $\mathbf Z[X]$, then for each $n$, the base-change $Y_n$ of $Y$ under the map $$\mathbf Z[X] \to \mathbf Z : X \mapsto n$$ would be finite étale over $\mathbf Z$, hence trivial (as you surely know that $\pi(\mathbf Z) = 0$). It seems that the coefficients of a polynomial defining $Y$ would have to vanish at every integer...

It's no reason that the étale fundamental group is hard or even impossible to calculate. Absolute Galois groups, the simplest case, are very difficult to calculate! (What does it even mean to "calculate" the absolute Galois group of $\mathbf Q$? I don't know!)

Another nice example comes from elliptic curves. If $E/\mathbf Q$ is an elliptic curve, then $\pi(E) = TE,$ the 'global' Tate module of $E$ : $$TE = \varprojlim_n E[n],$$ where the integers are ordered by divisibility ($E[n]$ denotes the $n$-torsion of $E$ over $\overline{\mathbf Q}$). It is a free $\widehat{\mathbf Z}$-module of rank $2$ (note that it's abelian, a rare feat for a fundamental group!). This is the "$\text{GL}_2$" analogue of the fact (which you probably also know) that $\pi(\mathbf G_m) = \widehat{\mathbf Z}$, where $\mathbf G_m = \text{Spec }\mathbf Q[X, X^{-1}]$ is the multiplicative group over $\mathbf Q$. The proof goes as follows: if $f: E' \to E$ is a finite étale covering of degree $n$, then $E'$ must have genus $1$ by Riemann-Hurwiz. By possibly extending the base field, it acquires a point, in such a way that $f$ becomes an isogeny. By pre-composing $f$ with the dual isogeny, we get the map $E \to E' \to E$ which is just multiplication by $n$ on $E$ (which is defined over $\mathbf Q$!). Thus the multiplication-by-$n$ maps form a cofinal system in the category of finite étale coverings, and the result follows.

Also, it is no coincidence that the fundamental group (in the usual sense) of a torus is $\mathbf Z^2$!