Here is a nice conceptual way which makes the continuity much clearer. It may not be in the form that you want, but you can certainly use it to prove the result in your language.
So, let's assume that $f:X\to\mathrm{Spec}(\mathbb{Q})$ is smooth proper (as we always do). We then have the following nice fact:
Theorem(étale Ehrassman's theorem): Let $f:X\to Y$ be smooth proper. For any LCC sheaf $\mathcal{F}$ on $X$, the sheaves $R^if_\ast\mathcal{F}$ are LCC on $Y$.
In particular, note that for our $f:X\to\mathrm{Spec}(\mathbb{Q})$ we have that $R^if_\ast\underline{\mathbb{Z}/\ell^n\mathbb{Z}}$ is an LCC sheaf on $\mathrm{Spec}(\mathbb{Q})$. But, by standard theory, this is equivalent to giving a continuous, finite $G_\mathbb{Q}$-module $M_n$. In particular, since the obvious compatibilities hold, one sees that
$$M:=\varprojlim M_n=\varprojlim R^if_\ast\underline{\mathbb{Z}/\ell^n\mathbb{Z}}$$
is a continuous $\mathbb{Z}_\ell$-representation of $G_\mathbb{Q}$.
Why does this matter to us? Well, by smooth proper base change, if $\overline{x}:\mathrm{Spec}(\overline{\mathbb{Q}})\to\mathrm{Spec}(\mathbb{Q})$ is any geometric point, then
$$(R^if_\ast\underline{\mathbb{Z}})_\overline{x}=H^i_\mathrm{\acute{e}t}(X_{\overline{\mathbb{Q}}},\mathbb{Z}/\ell^n\mathbb{Z})$$
But, taking stalks is precisely the equivalence
$$\left\{\begin{matrix}\text{LCC sheaves}\\\text{on }\mathrm{Spec}(\mathbb{Q}\end{matrix}\right\}\longleftrightarrow\left\{\begin{matrix}\text{finite continuous}\\G_\mathbb{Q}\text{-modules}\end{matrix}\right\}$$
Thus, you see that $M_n=H^i_{\mathrm{\acute{e}t}}(X_\overline{\mathbb{Q}},\mathbb{Z}/\ell^n\mathbb{Z})$ as $G_\mathbb{Q}$-modules, from where continuity follows. And, of course, $M=H^i_{\mathrm{\acute{e}t}}(X_{\overline{\mathbb{Q}}},\mathbb{Z}_\ell)$ as $G_\mathbb{Q}$-modules.