Noah Schweber already gave a satisfactory answer, but let me make an extended comment that also gives an answer (not in the most efficient way, but in a way that is, I hope, both instructive and "constructive" to some extent), with a computer implementation of sorts.
I think that I'm somewhat responsible for the Internet popularity of the "matchstick" representation of ordinals when I uploaded this image of $\omega^2$ to Wikipedia which has, since, crept up all over the place (e.g., in Joel David Hamkins's answer in this question), even though much better images have been produced afterwards.
(Incidentally, I don't remember where I learned about this matchstick representation in the first place. I know it appears in Paul Taylor's book Practical Foundations of Mathematics, and I think that's where I got the name "matchstick" to use on Wikipedia, but I think I had seen similar diagrams much earlier… Perhaps in Gamow's One Two Three Infinity? I don't have it anymore so I can't check.)
It's easy to show that every countable ordinal is order-isomorphic to a closed subset of $\mathbb{R}$ (or a to a discrete one, if we prefer), but perhaps we want stronger constraints, like placing the matchsticks at rational or even dyadic positions and/or imposing regularity constraints like you do.
(For instance, I spent a lot of time asking myself whether it was more æsthetically appealing to place the matchsticks representing $\omega$ at exponentially decreasing intervals like $\sum_{k=1}^N\frac{1}{2^k}$, or instead as something like $1-\frac{1}{N}$ which is how one would see the rungs of an infinite ladder in perspective. I ended up settling on the former, but the latter may be intellectually more satisfactory.)
Anyway, in trying to systematize this "matchstick" representation of ordinals, I wrote this interactive page which uses JavaScript (and HTML5 canvas) to display various ordinals, which you can then click on various parts in order to zoom. I would like to say a word about how it works. The JavaScript code is readable (it's not obfuscated or minified, just choose "view source" and you can read it inside the HTML), and one of the interesting features is that I coded an ordinal representation system for ordinals up to $\zeta_0 := \varphi(2,0)$ (the smallest fixed point of $\alpha \mapsto \varphi(1,\alpha) := \varepsilon_\alpha$) and used it to systematically display names and matchstick representations of the ordinals.
Two things are missing for the program in question to answer your question: (1) you ask for the position of the matchsticks to be dyadic rationals and for the ratio of distances between two successive matchsticks to be $1/2$, and (2) you want the ordinals represented to be much greater than $\zeta_0$. Now (1) is trivially fixed, in fact, there is a variable called magicProp in the code which, if set to 0.5 (instead of the default 0.3), will give exactly the condition you ask for. As for (2), it is clear that $\zeta_0$ plays no particular rôle here: it just so happens that I was too lazy to code a representation system for much larger ordinals in JavaScript, but it can be done and plugged into the program; obviously any computer program can only deal with ordinals less than $\omega_1^{\mathrm{CK}}$, but you did not ask for a computer program.
So, to wrap it all up, here is one simple lemma which basically makes my program work and solves your question:
Lemma: Any ordinal $\alpha < \omega_1$ which is of the form $\omega^\gamma$ with $\gamma>0$ can be written as the sum of a sequence $\sum_{i<\omega} \beta_i$ where each $\beta_i < \alpha$ is itself a power of $\omega$.
Proof. If $\gamma = \gamma'+1$ is successor, write it as the sum of $\omega$ terms all equal to $\omega^{\gamma'}$. Otherwise, write $\gamma$ as limit of an increasing sequence $(\gamma_i)_{i<\omega}$, then $\omega^\gamma = \sum_{i<\omega} \omega^{\gamma_i}$ (because in fact $\sum_{i\leq n} \omega^{\gamma_i} = \omega^{\gamma_n}$), as announced.
(In my JavaScript program, this lemma is implemented by the function summingSequence, essentially with the proof I just gave.)
Now inductively define a matchstick ordinal representation inside $[0,1)$ for any $\alpha < \omega_1$ which is a power of $\omega$ as follows: if $\alpha = 1$ just use a single matchstick at position $0$; otherwise, write $\alpha = \sum_{i<\omega} \beta_i$ as given by the lemma, divide the interval $[0,1)$ into countably many consecutive subintervals, each one $p$ times the length of the previous one (for what you ask, take $p=1/2$, whereas my program uses $p=3/10$), and place a scaled copy of the matchstick representation of each $\beta_i$ inside each consecutive interval.
You asked for the interval between the two first matchsticks to be $1$, but this is immediately solved by scaling the above construction (simply note that if $p=1/2$, the interval it provides is of the form $1/2^k$ for some $k$). The case of ordinals which aren't pure powers of $\omega$ is easily handled by the Cantor normal form and I won't describe it.
Anyway, my answer is essentially the same as Joel David Hamkins's, but I wanted to emphasize that it can be more or less implemented (up to whatever ordinal you are willing to code an ordinal notation system for).
Of course, the disappointing fact is that such representations are not at all enlightening. When I first saw matchstick representations, I thought, oh, this is wonderful, if I could draw one for some difficult-to-imagine ordinal like $\varepsilon_0$ it would help me visualize $\varepsilon_0$, but of course this is not at all true, as the interactive JavaScript page shows: with the construction described above, all large ordinals look boringly the same, they are visually indistinguishable from the obvious self-similar set consisting of countably many copies of itself laid end to end (each $p$ times the size of the previous one).
Let me preface this by stating the obvious in that there is no injection from a given uncountable cardinal to any countable ordinal, thus any system of describing "analogous properties" of large countable ordinals relative to an uncountable set of ordinals will be flawed in infinitely many ways. The connection between large cardinal axioms and large unrecursive countable ordinal axioms is based on intuition, not that there is anything wrong with that. The bottom line is, "recursively greatly Mahlo" ordinals can all be stable, or nonprojectible or $\Pi_3$-reflecting if you define them as such, but they absolutely don't have to be.
I suggest reading "Generalizing the Mahlo hierarchy, with applications to the Mitchell models" by Baldwin, where he defines a very structured hierarchy of Mahloness in analogy to the Mitchel rank for measurability of cardinals. I myself have not read more than a few pages but we only need to use the outline from the paper, so the descriptions I will give here are very oversimplified and do not constitute definitions, so keep that in mind. $m(\kappa)$ is a rank of Mahloness that essentially tells the that a filter on the stationary subsets of $\kappa$, if applied less than $m(\kappa)$ times to an equivalence class on $\kappa$
based on stationarity, will result in a set that is neither 0 nor the set of all nonstationary subsets of $\kappa$. This ranking does not provide much new insight for the structure of Mahloness as $m(\kappa)\leq\kappa^+$ is always the case by definition and for every greatly Mahlo $\kappa$ we already have $m(\kappa)=\kappa^+$ and vice versa, thus this notation is useless for trying to understand cardinals that go beyond greatly Mahlos in stationary properties that are themselves not on the level of the hierarchy of weakly and supercompacts. Further down the paper in section 3, functions $\bar m(\kappa), m^*(\kappa)$ are defined with more complex definitions that provide further structure to the different types of Mahlo cardinals and the different levels of iterated stationarity. According to Baldwin, for these to work as desired, we must assume a particular type of failure of GCH in $2^\kappa=2^{\kappa^+}$ at least for greatly Mahlo $\kappa$, that is, $|\mathcal P(\kappa)|$, which is the size of the domain and codomain of $m(), \bar m(), m^*()$, must be greater or equal to a limit of regular cardinals above $\kappa$. There is no analogy of this that can be translated down to the constructible universe of countable sets $L_{\omega_1}$, which is where we want the least "recursively greatly Mahlo" to fall.
Let $M[\beta](0)$ be the first recursively $\beta$-Mahlo ordinal and let $M[\beta](\lambda+\alpha)$ be the $\alpha$th recursively $\beta$-Mahlo ordinal above $M[\beta](\lambda)$ whenever $\alpha < M[\beta+1](0)$ and let $\beta$-Mahlo ordinals be exactly those ordinals $\kappa$ such that $L_\kappa\models$KP+"$\forall\gamma<\beta:$ the $\gamma$-Mahlo ordinals form a stationary subset of the ordinals".
We could define a collapse of $M[\beta+1](\alpha)$ inside $M[\beta](\text{___})$ to form an evaluation for the hierarchy of $\beta$-Mahlos, limits of such, inaccessible limits of such, hyper-inaccessible limits of such, hyper-hyper-inaccessible limits of such and so on, and keep going from there. If we use, say, the least $\Pi_3$-reflecting ordinal $K$ to collapse inside of $M[\text{___}](0)$ such that $M[K+\gamma](\beta)$ is the $(1+\beta)$th ordinal $\alpha$ that is $\alpha$-Mahlo and for all $\delta_0\in\gamma\land\delta_1\in\alpha:$ the ordinals of the form $M[K+\delta_0](\delta_1)$ form a stationary subset of $\alpha$ with respect to clubs definable in $L_{\omega_1}$. We can have an ordinal representation system $M[K+K\alpha+\beta](\gamma)$ being the $(1+\gamma)$th $\beta$-hyper$^\alpha$-Mahlo ordinal and so on. Then finally, we can define the recursively greatly Mahlo ordinals as exactly those that are of the form $M[\omega^{\rm CK}_{K+1}](\alpha)$ for some $\alpha$ as $\omega^{\rm CK}_{K+1}$ is admissible and above $K$ and thus not reacheable from below by some recursive well-ordering on $K+1$ and thus the least recursively greatly Mahlo will be greater than the $M$-collapse of any ordinal in $[K,K^+)$ or even simply in $K^+$.
If $\alpha^+$ is the least regular above $\alpha$, that means that there is no surjection from $\alpha$ to $\alpha^+$. Analogously, if $\alpha^+$ is the least admissible above $\alpha$ then that means there is no surjection from $\alpha$ to $\alpha^+$ that consitutes a recursive or hyperarithmetical reordering of $\alpha$. This is not a definition but it suffices for making the analogy.
I believe this way of defining the recursively greatly Mahlos fits the bill for placing them in the hierarchy of recursively Mahlos in such a way that they always have the recursively $\beta$-hyper$^\alpha$-Mahlo ordinals and all kinds of limits of such be unbounded in them, yet without the greatly Mahlos necessarily having $\Pi_3$-reflecting or stronger properties.
Best Answer
As per the many comments, the issue is confusing ordinal and cardinal arithmetic. The number of such expressions that yield countable ordinals is uncountable, as it should be. This is because the $\beta$ can range over all countable ordinals (of which there $\aleph_1$ many). Thank you to all commenters for replying so quickly.