Definable order types without infinity axiom.

To move this off the unanswered queue (actually user 喻良 already posted the correct answer but it was deleted as a link-only answer; also, it only addressed the first question):

The answer is indeed $\omega_1^{CK}$, the least noncomputable ordinal. To see this it's convenient to switch from $V_\omega$ to $\mathbb{N}$; they are bi-interpretable (via e.g. the Ackermann interpretation), so we can basically always switch from one to the other.

By definition, $\omega_1^{CK}$ is the smallest ordinal with no computable copy - that is, the least ordinal $\alpha$ such that there is no computable relation $R$ on $\omega$ such that $R$ defines a well-ordering of $\omega$ of type $\alpha$. Since computable = $\Delta^0_1$, every ordinal with a computable copy in fact has a definable copy. So your ordinal is $\ge\omega_1^{CK}$. But in fact $\omega_1^{CK}$ is gigantic: it is also the smallest ordinal with no hyperarithmetical copy. This is a theorem due to Spector - see Corollary $5.6$ of chapter $1$ of Sacks' book. Since $\mathbb{N}$-definable = arithmetical $\subsetneq$ hyperarithmetical, this shows that your ordinal is $\le\omega_1^{CK}$.

What about bringing classes into the picture? Well, by implementing the $L$-construction in the setting of second-order arithmetic (see the last section of Simpson's reverse math book) this gets us all the reals in $L_\alpha$ for $\alpha$ the smallest "gap ordinal," that is, the smallest $\alpha$ such that $L_{\alpha+1}\cap\mathbb{R}=L_\alpha\cap\mathbb{R}$. This ordinal is absolutely gigantic, certainly vastly bigger than $\omega_1^{CK}$. So yes, things increase at that point.

That's such a big jump that it feels a bit rude. A gentler hike up the countable ordinals goes through looking at the smallest ordinal with no copy definable over $L_\alpha$ for appropriate $\alpha$. This is always at most the next admissibile above $\alpha$, and can be strictly smaller - indeed it usually is, in the sense that this inequality is strict on a club. This is proved similarly to here. One important issue to keep in mind is that, contra the $\omega$-setting, "definable over definable $\not=$ definable" for ordinals over general admissible sets. The robustness of $\omega_1^{CK}$ is rather special.

Incidentally, there's a good "atlas" of countable ordinals here.