All countable ordinals are embeddable in $\mathbb{Q}$.
For "small" countable ordinals, it is simple to do this explicitly.
$\omega$ is trivial, $\omega+1$ can be e.g. done as $\{\frac{n}{n+1}:n\in \mathbb{N}\} \cup \{1\}$.
$\omega*2$ can be done as $\{\frac{n}{n+1}:n\in \mathbb{N}\} \cup \{1+\frac{n}{n+1}:n\in \mathbb{N}\}$, and $\omega*n$ as $\bigcup_{i\le n} \{i+\frac{n}{n+1}:n\in \mathbb{N}\}$
That immediately also gives $\omega^2$ as $\bigcup_{i \in \mathbb{N}} \{i+\frac{n}{n+1}:n\in \mathbb{N}\}$
And going further is still relatively easy: We can biject the above embedding of $\omega^2$ onto a single interval $(0,1)$, e.g. through $f(x)=1-\frac{1}{x+1}$ since this is an order preserving bijection from $\mathbb{Q}^+$ to $(0,1)$, allowing us to get $\omega^2+n$, $\omega^2*n$ etc. And by iterating that process, we can get any ordinal below $\omega^\omega$.
But this sort of embedding fails at $\omega^\omega$ as the iteration of $f$ doesn't seem compatible with taking the infinite union – figuratively speaking, we'd squish things to $0$. So what would an explicit embedding of $\omega^\omega$ look like? And the question, then, is if it is possible to give an explicit embedding of $\epsilon_0$? Is the fact that Peano arithmetic cannot prove well-foundedness of $\epsilon_0$ an indication that it cannot be embedded by such elementary functions and their iterations?
And what about even bigger countable ordinals such as the Veblen ordinals, Feferman-Schütte, Bachmann-Howard? From its definition, even if the above all are possible, I assume that the definitive bound of where one can define an effective procedure for the embedding must be Church–Kleene – is that a correct conclusion?
Lastly, does the situation change when we use $\mathbb{R}$ instead of $\mathbb{Q}$, which would allow us to use, perhaps, more easily understood yet inherently complex functions?
Edited because $\omega^2 \neq \omega^\omega$
Best Answer
Since $\epsilon_0$ is still an $\omega$-sequence, it seems to me that an embedding is relatively straightforward; take it as the sequence $\left\{\omega, \omega^\omega, \omega^{\omega^\omega}, \ldots\right\}$ and embed each of those in a unit interval. Each has an easy explicit embedding, and for any '$\omega$-polynomial' that you give me I can give you the rational in my embedding that corresponds to it. I suspect that things break down, as you say, not too many steps up in the hierarchy - but that starts to get to questions of what an 'explicit specification' is.
EDIT: And to answer the question about embedding $\omega^\omega$, the same can easily be done; to make it more straightforward and highlight my mapping above, I'll pack it into the unit interval $\left[1,2\right)$. Map $\omega$ onto $\left[1,1+\frac{1}{2}\right)$, $\omega^2$ onto $\left[1+\frac{1}{2}, 1+\frac{3}{4}\right)$, etc; then our effective procedure for 'decoding' a polynomial $\sum_{i=0}^na_i\omega^i$ to a rational produces the rational $1+(1-2^{-n})+2^{-(n+1)}\left(1-2^{-(a_n-1)}\right)+2^{-(n+1)}2^{-a_n}\left(1-2^{-(a_{n-1}-1)}\right)+\cdots$ — we 'chop off' the largest term so that we're working in the interval $\left[1+(1-2^{-n}), 1+(1-2^{-(n+1)})\right)$ of length $2^{-(n+1)}$, use $a_n$ to find the point in that interval representing $a_n\omega^n$ (and implicitly, the next 'mapped-down' interval), and repeat the procedure. As long as there's an explicit means of specifying an $\omega$-sequence for the ordinal, this general mechanism will work.