Hopefully this question is well defined. Consider the following linear order in the language $\{<\}$:
Step 0: Begin with $\mathbb{Q}$.
Step 1: Create a new model $Q_1$ by realizing all the types (cuts) in $S_1^\mathbb{Q}([-1,1])$.
Step 2: Create a new model $Q_2$ by realizing all the types (cuts) in $S^{Q_1}_1([-\frac{1}{2},\frac{1}{2}])$.
Step $n$: Create a new model $Q_{n}$ by realizing all the types (cuts) in $S^{Q_{n-1}}_1([-\frac{1}{n},\frac{1}{n}])$.
Now let $Q_\omega =\bigcup_{n<\omega}Q_n$.
Questions: Does $|S^{Q_\omega}_1(Q_\omega)|= \beth_\omega$? If GCH holds, then does $|S^{Q_\omega}_1(Q_\omega)|=|Q_\omega|$ and $Q_\omega$ is a stable dense linear order without endpoints? Is the stability of this model independent of ZFC?
Thanks.
If I understand your construction correctly, I don't see that $Q_\omega$ can have a cardinality higher than $\mathfrak c$. More precisely, each $x\in Q_\omega$ can be described by a finite sequence of reals as follows:
If $x$ was added in step $n$ as the "principal" infintesimal right above or below an element of $y$ that already existed, describe it as $\langle n,\pm 1\rangle$ followed by the description of $y$.
If $x$ was added in step $n$ as part of the "buffer copy" of $\mathbb Q$ betewen an already existing element $y$ and the infinitesimal above or below it, describe it as $\langle n,\pm2,q\rangle $ followed by the description of $y$, where $q$ is the location within the buffer copy.
Otherwise, $x$ was added in step $n$ corresponding to an irrational cut in one of the "buffer copies" of $\mathbb Q$ from step $n-1$. Represent that as $\langle n-1,\pm2,\alpha\rangle$ where $\alpha$ is the irrational cut, and followed by the description of of the same $y$ that was used to describe the insertion of the buffer copy itself.
(Once each of the cuts in each copy of $\mathbb Q$ has been filled in the next step, the subsequent steps already has that cut filled; it doesn't need to be replicated in subsequent steps. And every "cut of cuts" determines a cut we already know, either from above or below; that's the point of the reals being a complete order).
Thus $|Q_\omega|=\beth_1$, and $S_1^{Q_\omega}(Q_\omega)$ can't, as a matter of syntax, be larger than $\beth_3$ or thereabouts -- actually, with coding similar to those above, I think $S_1^{Q_\omega}(Q_\omega)$ ifself has cardinality $\mathfrak c$. In any case it's a far cry from $\beth_\omega$.