Here's a nice way to see that $\mathrm{Spec}(\mathbb{Z})$ is étale simply connected. Since $\mathrm{Spec}(\mathbb{Z})$ is a normal scheme every connected finite étale cover of $\mathrm{Spec}(\mathbb{Z})$ is of the form $\mathrm{Spec}(\mathcal{O}_K)$ for some number field $K$, as $\mathcal{O}_K$ is the normalization of $\mathbb{Z}$ in $K$ over $\mathbb{Q}$. Now suppose that $K$ has degree $n=r_1+2r_2$ where $r_1$ and $r_2$ are the numbers of real and complex embeddings respectively of $K$, and let $\Delta_K$ be the discriminant of $K$. Then by the Minkowski bound (see Neukirch) every class in the ideal class group of $K$ contains an integral ideal of norm at most the Minkowski constant $M_K=\sqrt{\vert\Delta_K\vert}(\frac{4}{\pi})^{r_2}\frac{n!}{n^n}$. But by definition an integral ideal of $\mathcal{O}_K$ has norm at least $1$ so that $1\leq M_K$ and hence $\sqrt{\vert\Delta_K\vert}\geq(\frac{\pi}{4})^{r_2}\frac{n^n}{n!}\geq(\frac{\pi}{4})^{n/2}\frac{n^n}{n!}$; in particular if $K$ is a nontrivial extension of $\mathbb{Q}$ we have $\vert\Delta_K\vert>1$, and since a prime $p\in\mathbb{Z}$ ramifies in $K$ precisely if $p$ divides $\Delta_K$ it follows that the finite étale cover $\mathrm{Spec}(\mathcal{O}_K)\rightarrow\mathrm{Spec}(\mathbb{Z})$ is somewhere unramified unless it is trivial, which implies $\widehat{\pi}_1(\mathrm{Spec}(\mathbb{Z}))=0$. 

In general for $\mathcal{O}_K$ the ring of integers of a number field $K$ the maximal Abelian quotient $\widehat{\pi}^\mathrm{ab}_1(\mathrm{Spec}(\mathcal{O}_K))$ is isomorphic to the narrow class group $\mathrm{Cl}^+_K=I_K/P^+_K\simeq\mathrm{Gal}(H^+_K/K)$ where where $H^+_K$ is the narrow class field of $K$ (the maximal Abelian extension of $K$ which is unramified outside the finite places), and $I_K/P^+_K$ is the group of fractional ideals of $\mathcal{O}_K$ modulo the group of principal fractional ideals $(\alpha)=\alpha\mathcal{O}_K$ whose generator $\alpha\in K$ is totally positive for each real embedding.