Expanding a ring doesn't always mean shrinking the associated some schemes. However, expanding a ring by specifically adding inverses of elements that didn't already have inverses will shrink the Spec space. The inclusion $\Bbb Z\to\Bbb Z[1/n]$ will induce an inclusion $\operatorname{Spec}\Bbb Z[1/n]\subseteq \operatorname{Spec}\Bbb Z$.
In contrast, $\operatorname{Spec}\Bbb Z$ is well-known and easy to describe, while $\operatorname{Spec}\Bbb Z[X]$ is a complicated beast, and I don't think we are anywhere near a description of $\operatorname{Spec}\Bbb Z[X,Y]$.
If you have two rings $R, S$ and a ring homomorphism $R\to S$, then that induces a morphism of affine schemes $\operatorname{Spec}S\to\operatorname{Spec}R$. And if $X$ is a $\operatorname{Spec}S$ scheme, that just means there is a morphism $X\to\operatorname{Spec}S$. Composing that with the morphism $\operatorname{Spec}S\to\operatorname{Spec}R$ gives you a morphism $X\to \operatorname{Spec}R$, which is to say that $X$ is indeed a scheme over $\operatorname{Spec}R$. This is true for $R = \Bbb Z$ and $S = \Bbb Z[1/n]$ in particular.
There is a morphism from any scheme $X$ to $\operatorname{Spec}\Bbb Z$. It takes each point in $X$ to the (prime ideal generated by the) characteristic of the residue field over that point. So any scheme is a $\Bbb Z$-scheme.
The idea of taking limits is the right one.
Let $A = \overline{\mathbb{Q}}\otimes_{\mathbb{Q}}\overline{\mathbb{Q}}$.
Fact 1. All prime ideals of $A$ are maximal and all its residue fields are canonically isomorphic with $\overline{\mathbb{Q}}$.
Proof. For the proof note that $\mathbb{Q}\hookrightarrow \overline{\mathbb{Q}}$ is an integral morphism. Hence $\overline{\mathbb{Q}} \hookrightarrow A$ is an integral morphism (as a base change of the previously considered morphism). Pick a prime ideal $\mathfrak{p}\subseteq A$, then the induced map $\overline{\mathbb{Q}} \hookrightarrow A/\mathfrak{p}$ is integral. Now if $K\hookrightarrow D$ is an integral morphism, where $K$ is a field and $D$ is an integral domain, then $D$ is a field. Thus $A/\mathfrak{p}$ is a field algebraic over $\overline{\mathbb{Q}}$. Hence $\mathfrak{p}$ is maximal and $A/\mathfrak{p} \cong \overline{\mathbb{Q}}$.
Fact 2. Let $R$ be a commutative ring such that all prime ideals of $R$ are maximal. Then $\mathrm{Spec}\,R$ is a Hausdorff topological space. In particular, $\mathrm{Spec}\,R$ is compact.
Proof. By dividing $R$ by nilradical, we may assume that $R$ has no nilpotent elements. Since every prime ideal $\mathfrak{p}$ of $R$ is maximal, we deduce that $R_{\mathfrak{p}}$ is a reduced local ring with single prime ideal. Hence $R_{\mathfrak{p}}$ is a field. Pick now distinct prime ideals $\mathfrak{p}$, $\mathfrak{q}$ of $R$. Since they are maximal, there exists $f\in \mathfrak{p}\setminus \mathfrak{q}$. Now $\frac{f}{1}\in R_{\mathfrak{p}}$ is zero (ideal $\mathfrak{p}R_{\mathfrak{p}}\subseteq R_{\mathfrak{p}}$ is zero as $R_{\mathfrak{p}}$ is a field). Thus there exists $g\not \in \mathfrak{p}$ such that $g\cdot f =0$. Since $g\cdot f = 0 \in \mathfrak{q}$ and $f\not \in \mathfrak{q}$, we deduce that $g\in \mathfrak{q}$. Thus $g\in \mathfrak{q}\setminus \mathfrak{p}$ and $f\cdot g = 0$. Therefore, we have (for distinguished open subsets of $\mathrm{Spec}\,R$)
$$\mathfrak{p}\in D(g),\,\mathfrak{q}\in D(f),\,D(f)\cap D(g) = \emptyset$$
and thus $\mathrm{Spec}\,R$ is a Hausdorff space. Since it is quasi-compact, we deduce that it is compact.
Now we are ready to prove that $\mathrm{Spec}\,A$ is homeomorphic to $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$. Let $\mathcal{K}$ be a diagram consisting of all finite extensions of $\mathbb{Q}$ and inclusions between them. Consider the limit $X$ in the category of topological spaces of the induced diagram
$$\big\{\mathrm{Spec}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)\big\}_{K\in \mathcal{K}}$$
Clearly $X$ is homeomorphic with $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ (this is just the definiton of a profinite topology on a Galois group). It suffices to check that $X$ is homeomorphic with $\mathrm{Spec}\,A$. Note that
$$A=\mathrm{colim}_{K\in \mathcal{K}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)$$
in the category of $\overline{\mathbb{Q}}$-algebras. Hence $\mathrm{Spec}\,A$ gives rise to a cone over topological diagram $\big\{\mathrm{Spec}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right)\big\}_{K\in \mathcal{K}}$ and thus there exists a unique comparison map $f:\mathrm{Spec}\,A\rightarrow X$. Clearly $f$ is continuous. We have
$$\mbox{Points of }\mathrm{Spec}\,A\,=\mathrm{Mor}_{\overline{\mathbb{Q}}}\left(A, \overline{\mathbb{Q}}\right) = \mathrm{Mor}_{\overline{\mathbb{Q}}}\left(\mathrm{colim}_{K\in \mathcal{K}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K\right), \overline{\mathbb{Q}}\right) =$$
$$= \lim_{K\in \mathcal{K}}\mathrm{Mor}_{\overline{\mathbb{Q}}}\left(\overline{\mathbb{Q}}\otimes_{\mathbb{Q}}K, \overline{\mathbb{Q}}\right) = \mbox{Points of }X$$
where the first identification is a consequence of Fact 1. Moreover, (as you can check carefully) this identification is induced by $f$. Thus $f$ is bijective. Therefore, $f$ is a continuous bijection with compact source (by Fact 2) and Hausdorff target. Hence $f$ is a homeomorphism.
Best Answer
I'm just posting an answer to close this question.
Martin Brandenburg solved it and the result that he cites, that $F$ preserves filtered limits of affine morphisms, is 8.2.9 in EGA IV.3.