$\newcommand{\bomega}{\boldsymbol\omega}$Given the definition of bounding ordinal in the post and the potential sensitivity to coding mentioned in edit 2, these seem to be two main ways to formalize boundedness:
- $\gamma$ is bounding if for all $\alpha<\gamma$, for all ordinal notations $\{e\}$ computing $\alpha$, we have $\vert\mathsf{ACA}_0+``\{e\}\text{ computes a well-order}"\vert_{sup}\leq\gamma$.
- $\gamma$ is bounding if for all $\alpha<\gamma$, there exists an ordinal notation $\{e\}$ computing $\alpha$ such that we have $\vert\mathsf{ACA}_0+``\{e\}\text{ computes a well-order}"\vert_{sup}\leq\gamma$.
Edit Jun 29: By Hanul Jeon's answer, these are equivalent.
(Here an ordinal notation is a computable function computing the composition of the indicator function of a well-ordering with an unpairing function.) Taking either definition of boundingness, a result is possible which is relevant to the question in your edit: all bounding ordinals are epsilon ordinals.
Write $\prec_{\{e\}}$ for the well-ordering computed by a computable function $\{e\}$, $\mathrm{ot}(\prec)$ for the order type of a well-ordering $\prec$, and write $\mathrm{fld}(\prec)$ for the union of $\prec$'s domain and range.
The core of the following argument is Girard's equivalence result, and then three corollaries quickly follow.
Lemma: Let $T$ be a theory containing $\mathsf{ACA}_0$ and $\alpha$ be an ordinal. If there is an ordinal notation $\{e\}$ with order type $\alpha$ such that $T$ proves $\{e\}$ computes a well-order, then $\vert T\vert_{sup}\geq\omega^\alpha$.
Proof: Let $T$, $\alpha$, and $\{e\}$ be as given. Girard showed that arithmetical comprehension is equivalent over $\mathsf{RCA}_0$ to "for all linear orders $\prec$, if $\prec$ is well-founded then $\bomega^\prec$ is well-founded". (Marcone, Montalbán, "The Veblen functions for computability theorists", 2010). Here $\bomega^\prec$ is a kind of linear order exponentiation, briefly $\bomega^\prec$ is the lexicographic ordering on monotonically-$\prec$-decreasing finite tuples of members of $\mathrm{fld}(\prec)$. (The definition with detail is on page 4 of the previous source.)
Since $\prec_{\{e\}}$ is computable and $\bomega^{\prec_{\{e\}}}$ is essentially a lexicographic ordering on a computable set, $\bomega^{\prec_{\{e\}}}$ is computable. By looking at the Cantor normal form of $\mathrm{ot}(\prec_{\{e\}})$, the order type of $\bomega^{\prec_{\{e\}}}$ is $\omega^{\mathrm{ot}(\prec_{\{e\}})}$, and since $\mathrm{ot}(\prec_{\{e\}})=\alpha$, we have $\mathrm{ot}(\bomega^{\prec_{\{e\}}})=\omega^\alpha$. Let $\{d\}$ be an ordinal notation that computes $\bomega^{\prec_{\{e\}}}$. Thus we have an ordinal notation $\{d\}$ that computes the ordinal $\omega^\alpha$, and $T$ proves that it computes a well-order. Thus $\vert T\vert_{sup}\geq\omega^\alpha$. $\square$
Corollary 1: If $T$ contains $\mathsf{ACA}_0$, its PTO is an epsilon ordinal.
Proof: Let $\gamma=\vert T\vert_{sup}$, and $\alpha<\gamma$ be an arbitary ordinal. Let $\{e\}$ be an arbitrary ordinal notation of order type $\alpha$ that $T$ proves computes a well-order. By the lemma, we have $\vert T\vert_{sup}\geq\omega^\alpha$, i.e. $\gamma\geq\omega^\alpha$. As $\alpha<\gamma$ was arbitrary, $\gamma$ is an epsilon ordinal. $\square$
Corollary 2: Any bounding ordinal (using the first definition) is an epsilon ordinal.
Proof: Assume $\gamma$ is a bounding ordinal, using the first definition. Choose an arbitrary ordinal $\alpha<\gamma$ and an arbitrary ordinal notation $\{e\}$ computing $\alpha$. Then $\vert\mathsf{ACA}_0+``\{e\}\text{ computes a well-order}"\vert_{sup}\geq\omega^\alpha$, and by boundingness this ordinal is $\leq\gamma$. As $\alpha<\gamma$ was arbitrary, we have shown that for any $\alpha<\gamma$ we have $\omega^\alpha\leq\gamma$, so $\gamma$ must be an epsilon ordinal. $\square$
Corollary 3: Any bounding ordinal (using the second definition) is an epsilon ordinal.
Proof: Assume $\gamma$ is a bounding ordinal, using the second definition. Choose an arbitrary ordinal $\alpha<\gamma$ and an ordinal notation $\{e\}$ computing $\alpha$ such that $\vert\mathsf{ACA}_0+\{e\}\text{ computes a well-order}\vert_{sup}\leq\gamma$. Then $\vert\mathsf{ACA}_0+``\{e\}\text{ computes a well-order}"\vert_{sup}\geq\omega^\alpha$, and by boundingness this ordinal is $\leq\gamma$. As $\alpha<\gamma$ was arbitrary, we have shown that for any $\alpha<\gamma$ we have $\omega^\alpha\leq\gamma$, so $\gamma$ must be an epsilon ordinal. $\square$
I am not sure how to show that any particular ordinal greater than $\varepsilon_0$ is limiting using these definitions, or even if $\varepsilon_0$ is limiting using the first definition. For these problems it suffices to show that there is no pathological computable well-ordering $\prec$ with $\mathrm{ot}(\prec)$ less than a desired $\gamma$ such that well-foundedness of $\prec$ implies well-foundedness of a computable well-ordering $\prec'$ with much higher order type. I have been unable to produce such counterexamples well-ordering $\prec$, and also unable to prove one does not exist. Maybe something in the spirit of the "Strong statements from small ordinals" paragraph of Dmytro Taranovsky's answer to MO question #432470 would help with constructing such a $\prec$, but I am unable to extend Taranovsky's work all the way up to $\Pi^1_1$ sentences, like "$\prec'$ is a well-ordering". I have also considered the Kleene-Brouwer ordering on a tree searching for an infinite decreasing sequence in $\prec'$, but was unable to show that it would have small order type.