The homotopy groups of the étale topos of a scheme were defined by Artin and Mazur.  Are these known for Spec Z?  Certainly π1 is trivial because Spec Z has no unramified étale covers, but what is known about the higher homotopy groups?
$Spec(\mathbb{Z})$ should only be considered as $S^3$, if you "compactify" that is add the point at the real place. This is demonstrated by taking cohomology with compact support.
The étale homotopy type of $Spec(\mathbb{Z})$ is however contractible (indeed what do you get by removing a point form a sphere?)  to see this (all results apper in Milne's Arithmetic Dualities Book)
(let $X=Spec(\mathbb{Z})$ )

$H^r_c(X_{fl},\mathbb{G}_m)=H^r_{c}(X_{et},\mathbb{G}_m) = 0$ for $r \neq 3$.

$H^3_c(X_{fl},\mathbb{G}_m)=H^3_{c}(X_{et},\mathbb{G}_m) = \mathbb{Q}/\mathbb{Z}$

by 2+1, we have:


$H^3_c(X_{fl},\mu_n)= \mathbb{Z}/n$
$H^r_c(X_{fl},\mu_n)= 0$ for $r \neq 3$.

since we have a duality $$H^r(X_{fl},\mathbb{Z}/n)\times  H^{3-r}_c(X_{fl},\mu_n) \to \mathbb{Q}/\mathbb{Z} $$
we have

$H^0(X_{fl},\mathbb{Z}/n) = H^0(X_{et},\mathbb{Z}/n) = \mathbb{Z}/n,$


$H^r(X_{fl},\mathbb{Z}/n) = H^r(X_{et},\mathbb{Z}/n) = 0$, $r >0$

Now  since $\pi_1$ is trivial we have by the Universal Coefficient Theorem, the Hurewicz Theorem and the profiniteness theorem for 'etale homotopy that all homotopy groups are zero.

