I think that the answer you want is "yes, once you make it rigorous". For example, if we are talking about theories of second-order arithmetic then it is clear what "T proves that index $e$ computes a well ordering" will mean. For theories of first-order arithmetic, the definition is more subtle, but for example it might mean that $T$ proves a certain axiom scheme that represents transfinite induction up to $e$, or $T$ proves some other axiom scheme.
In any case, the following will apply as long as "T proves that index $e$ computes a well ordering" means that $T$ proves some (possibly infinite) set of sentences that is uniformly r.e. given the index $e$; we will call the set $S(e)$.
There is another caveat that I am more nervous about. You are talking about arbitrary indices for computable well orderings. Usually, when we do ordinal analysis, we look at particular "nice" indices, which compute the well order in a nice way. It is known that there are various tricks that one can do with indices to hide what they are doing. For example, there is an index $e$ that really computes an order of type $\omega$, but Peano Arithmetic does not prove that the order computed by $e$ has a least element. So does that mean, in your setup, that the least ordinal that PA cannot prove to be well founded is $\omega$?
So, really, statements that claim to identify the proof-theoretic ordinal have to be understood as informal interpretations of particular results, rather than being statements about a formal definition of "proof theoretic ordinal". See https://mathoverflow.net/questions/52926/proof-theoretic-ordinal/52927#52927
I am going to ignore that temporarily, and give an answer in the spirit of the question, because it shows a key fact about $\mathcal{O}$. But I am not convinced that the thing that is being found is really "the proof theoretic ordinal of $T$".
Now I will proceed to the construction. Given an index $t$ for $T$, the statement "For every $i \in S(e)$, $T$ proves the sentence with Goedel number $i$" can be written as an arithmetical formula $\phi(t,e)$, where the same formula $\phi$ works for all $t$ and $e$. Basically, $\phi$ says: whenever $e$ enumerates a formula $i$, there is a formal proof of that formula from the set of axioms enumerated by $t$.
Thus there is also an arithmetical formula $\psi(t,e)$ which says that $e$ computes a linear order of $\omega$, and $\phi(t,e)$ does not hold but, for every $e' <_e e$, $\phi(t,e')$ does not hold. The last part quantifies over every proper initial segment of the ordering computed by $e$. Note that if $e$ actually computes a well ordering, then $\psi(t,e)$ holds if and only if the order type of $e$ is (in some imprecise sense, but I think the one you mean) the "proof theoretic ordinal of $T$."
The index $\mathcal{O}$ is very strong. In particular, given any arithmetical formula, we can use $\mathcal{O}$ to tell whether that formula is true or false.
Formally, this is the fact that $\emptyset^{(\omega)} <_T \text{HJ}(\emptyset) = \mathcal{O}$. Now, given $t$, we can do the following: enumerate all values of $e$. For each one, first ask whether $e \in \mathcal{O}$. If not, move on to the next value of $e$. If $e \in \mathcal{O}$, ask whether $\psi(e,t)$ holds. If so, we have found the ordinal for $t$. Otherwise, move on to the next value of $e$. This will find some sort of "proof theoretic ordinal" $e$, subject to my caveat above.