Your question as it stands is quite unclear, but let me take a stab at it; based on your previous questions, if nothing else I think you'll find this interesting.
The simplest interpretation of your question is to look for an analogue of the Busy Beaver function for iterates of the Turing jump through the ordinals. Such a thing does exist, although - as usual, and along similar lines to some of your previous questions - we have to index by ordinal notations instead of ordinals per se, and we run into trouble when we try to continue past the computable ordinals (see chapter $1$ of Sacks' book for the relevant background; on the other hand, see Hodes' paper on mastercodes for ideas on how to go even further):
For $l$ a notation for a computable limit ordinal, we get a sequence $(n_i)_{i\in\omega}$ of notations forming a canonical sequence for $\vert l\vert_\mathcal{O}$ (where "$\vert k\vert_\mathcal{O}$" denotes the ordinal $k$ is a notation for). The $l$th Busy Beaver function, $BB_l$, is then defined by
$$BB_l(x)=1+\sum_{i\le x}BB_{n_i}(x).$$
For $m$ a notation for a successor ordinal, we let $n$ be the canonical "predecessor notation" of $m$, and define $BB_m$ as the sum of $BB_n$ and the usual Busy Beaver function with oracle $BB_n$.
The $\mathcal{O}$-indexed "hierarchy" so defined has a number of nice properties. in particular, there is also a very tight connection with iterates of the Turing jump (as alluded to above): $BB_k$ is Turing-equivalent to $0^{(\vert k\vert_\mathcal{O})}$ for every notation $k$.
It turns out that this computation power carries over to domination: given any function $f$ dominating a Busy Beaver function with index a notation for a computable ordinal $\alpha$, we have $f\ge_T0^{(\alpha)}$.
Surprisingly, the degree $0^{(\alpha)}$ makes sense for $\alpha$ a computable ordinal, not just a notation - if $i,j$ are notations for computable ordinals, then the potentially-different sets $0^{(\vert i\vert_\mathcal{O})}$ and $0^{(\vert j\vert_\mathcal{O})}$ are of equal Turing degree.
This, in turn, lets us prove the desired domination result that $$\vert i\vert_\mathcal{O}<\vert j\vert_\mathcal{O}\implies BB_i\prec BB_j,$$ where "$\prec$" means "dominates: we argue by induction that for any notation $k$ the higher Busy Beaver function $BB_k$ dominates every function computable from $0^{(\alpha)}$ for any $\alpha<\vert k\vert_\mathcal{O}$.
This naturally suggests the question of what sets can be computed from fast-enough functions. A beautiful theorem of Solovay says that basically the situation above is the only way this can happen:
Given a set $A$, the following are equivalent: $(i)$ $A$ is hyperarithmetic, $(ii)$ there is some function $f$ such that every function dominating $f$ computes $A$.
The observation we've made about the $BB_i$s forms the $(i)$-to-$(ii)$ direction (the other direction requires a nice forcing argument). See Peter Gerdes' thesis for further work on the topic.
Yes, this is easy to do, and you've done it already.
Incidentally, I really think you should look at the first section of Sacks' book, which goes into detail about computable ordinals and which I think will clear up a lot of these issues.
Your guess is already one perfectly formal definition:
The minimal natural number $n$ such that there exists an $n$-state Turing machine that computes any ordinal $\alpha$ such that the proof-theoretical ordinal of $T$ is [smaller than] $\alpha$.
(Once we fix a precise meaning of "proof-theoretic ordinal," of course.) So I don't know what additional formality you want.
I've replaced "is embedded into" with "smaller than" since the former suggests that the embedding itself should be computable, which I'm not sure you want; I think the language I've used is less suggestive.
The key point is that the phrase "computes an ordinal" is perfectly precise: we mean that the Turing machine $\Phi$ in question computes the characteristic function of a binary relation $R$ on $\mathbb{N}$ such that $(\mathbb{N}; R)$ is a well-ordering. There are no additional assumptions on the Turing machine which need to be made in order for this to make sense (although of course not every Turing machine does compute an ordinal, and the set of indices of Turing machines computing ordinals is highly complicated$^1$).
That said, there are many variations you could consider:
You could replace "smaller than" with "actually equal to." That this is different in general shouldn't be surprising: think about how (in the context of $\mathbb{N}$) the Kolmogorove complexity of $a$ can be bigger than that of $b$, even if $a<b$.
You could focus on indices as opposed to states: that is, "... such that the $n$th computable well-ordering is ..."
You could ask for an embedding of the proof-theoretic ordinal into $\alpha$ to be "effective" in some sense. E.g. you could fix a computable copy $S$ of the proof-theoretic ordinal $\theta$ of $T$ (that is, $S$ is a computable binary relation on $\mathbb{N}$ such that $(\mathbb{N}; S)$ is isomorphic to the proof-theoretic ordinal of $T$), and then look for a pair of Turing machines, one of which computes a copy of some ordinal $\ge\theta$ and the other of which computes an embedding of $S$ into that copy.
And so forth.
However, all of these approaches merely "rename" the proof-theoretic ordinal; although natural numbers are simpler than ordinals, they're not any simpler (in this context) than computable ordinals, so it's not clear that this shift gains anything.
EDIT: This part is in response to Andres' comments below.
It's worth noting that the claim "the proof-theoretic ordinal of $T$ measures the strength of $T$" has problematic aspects.
(For simplicity, let's define the proof-theoretic ordinal of $T$, $pto(T)$, as the least ordinal having no "$T$-provable" computable copy; that is, the least ordinal $\alpha$ such that there is no $e$ such that $(i)$ the $e$th Turing machine computes a copy of $\alpha$ and $(ii)$ $T$ proves that the $e$th Turing machine computes a well-ordering. As long as $T$ is $\Pi^1_1$-sound, overspill implies that $pto(T)<\omega_1^{CK}$; conversely, it's not hard to show that if $T$ is not $\Pi^1_1$-sound then $pto(T)=\omega_1^{CK}$.)
The issue is:
What information does $pto(T)$ actually give us?
Remember that the proof-theoretic ordinal lives in a broader proof-theoretic context; "measuring the strength of $T$" is generally taken to mean much more than just having some "concrete" description of $pto(T)$ (whatever that means; note that we can trivially cook up a computable copy of $pto(T)$ - namely, $$\sum_{\mbox{$T\vdash$ "$\Phi_e$ is a well-ordering"}}\Phi_e$$ - so "concrete" here is a very loaded word!). For example, we want:
A good understanding of the computable functions which $T$ proves are total (e.g. for RCA$_0$ these are exactly the primitive recursive functions).
A natural system of notations leading up to $pto(T)$ (this is arguably folded into the word "concrete" above), together with a connection between these notations and proofs from $T$ (this is the new bit).
Some fixed notation for $pto(T)$ together with a proof from a very weak theory that the well-foundedness of the ordering corresponding to this notation implies the consistency of $T$ (this is closely related to the previous bulletpoint, and really should fall out of it directly).
And probably a bunch of other things proof theorists know about but I don't.
All of this goes beyond just finding a "concrete description" for $pto(T)$. (For example, how would we figure out anything about what computable functions PA proves are total just by being told out of thin air that $pto(PA)=\epsilon_0$?) So we have to be careful: especially when talking about strong theories (like ZF), we don't want to assume out of hand that $pto(T)$ is necessarily measuring as much as we want it to.
That said, I would say that $pto(T)$ measures the strength of $T$ (at least to some extent), and in this I might disagree with Andres below. But it is important (as he pointed out) to make this issue clear; $pto(T)$ isn't some magic box which tells us everything about the strength of $T$, and divorcing it from these ambient assumptions may make it (and by extension, questions like this) less mysterious.
$^1$I've linked to Kleene's $\mathcal{O}$ here; this isn't literally the set of indices of Turing machines computing ordinals, but it is in fact "equivalent" to that set (precisely: the two sets are Turing equivalent, indeed $1$-equivalent). The general theory of Kleene's $\mathcal{O}$, and of computable ordinals in general, is developed in the first part of Sacks' book.
Best Answer
Throughout, by "structure" I mean "countable structure in a computable language." I'm also assuming you're comfortable both with Turing reducibility $\le_T$ - which lets us avoid unnecessary verbiage about machines and oracles - and with the idea of reals coding copies of structures (see SSequence's answer, specifically the $\omega\cdot2$ example).
We begin with the computability side. For $r$ a real, we let $\omega_1^{CK}(r)$ be the smallest ordinal with no $r$-computable copy; equivalently, the supremum of the ordinals which do have $r$-computable copies. For a structure $\mathcal{A}$ we let $\omega_1^{CK}(\mathcal{A})$ be the smallest ordinal not computed by some$^1$ copy of $\mathcal{A}$; precisely, $$\omega_1^{CK}(\mathcal{A})=\min\{\omega_1^{CK}(r):r\mbox{ codes a copy of $\mathcal{A}$}\}.$$
Next, we look at the admissibility side. For $\alpha$ an arbitrary ordinal, we let $\omega_\alpha^{ad}$ denote the $\alpha$th admissible ordinal: that is, the $\alpha$th ordinal whose corresponding level of $L$ satisfies KP. Note that this definition has nothing to do with computability theory on the face of it (and in fact, doesn't even require $\alpha$ to be countable!). We'll also write "$\omega_1^{ad}(\beta)$" for the first admissible ordinal $>\beta$; in particular, $\omega_1^{ad}(\omega_\alpha^{ad})=\omega_{\alpha+1}^{ad}$.
Now Sacks' result (plus a bit of thought) shows that $$\omega_1^{CK}(\alpha)=\omega_1^{ad}(\alpha)\mbox{ for every countable ordinal $\alpha$}.$$ This is why you never see the "$ad$" notation: it's made completely irrelevant! In particular, "$\omega_\alpha^{CK}$" is just our "$\omega_\alpha^{ad}$."
Moreover, Sacks' result immediately implies that $\omega_1^{CK}(\mathcal{A})$, being the minimum of a set of admissible ordinals,is itself admissible.
Also, via forcing we can make sense of this for even uncountable $\alpha$. But that's really a side issue.
$^1$Note the careful quantification over copies here (and the implicit focus on "optimally simple" copies) in our definition of $\omega_1^{CK}(\mathcal{A})$. This is fundamental: different copies of the same structure can behave very differently, and we need to address this if our definitions are to be interesting at all.
Specifically, we can have very simple structures coded by very complicated reals: e.g. "swapping" $2n$ and $2n+1$ whenever $n\in 0'$ gives a copy of $\omega$ which computes $0'$, and more generally we can get copies of $\omega$ of arbitrarily high complexity. In fact, this (almost) always happens. So in order to say anything interesting, we need to talk about what all copies of a given structure can do.
What we're ultimately getting at here is the idea of reducibilities between structures. Here we're looking just at Muchnik (weak) reducibility: $\mathcal{A}\le_w\mathcal{B}$ if every real coding a copy of $\mathcal{B}$ computes some real coding a copy of $\mathcal{A}$. There are other reducibilities too - most immediately, Medvedev (strong) reducibility - but for these sorts of questions we're really in the Muchnik realm, at least for now.
EDIT: An important point here which I think will substantially clarify things is that Muchnik reducibility extends $\le$ - if $\mathcal{A}\ge_w\alpha$ and $\beta<\alpha$ then $\mathcal{A}\ge_w\beta$. In particular this means that $\omega_1^{CK}(\mathcal{A})$ is both the least ordinal without a copy computable from every copy of $\mathcal{A}$, and the supremum of the ordinals which do have copies computable from every copy of $\mathcal{A}$.
EDIT THE SECOND: And here's a way of constructing such a "sufficiently simple" copy of $\omega_1^{CK}$: a copy of $\omega_1^{CK}$ can be computed straightforwardly from Kleene's $\mathcal{O}$, but$^2$ $\mathcal{O}$ is in $L_{\omega_2^{CK}}$ and so every ordinal with a copy computable from $\mathcal{O}$ is $<\omega_2^{CK}$. All of this requires a bit of familiarity with admissible sets and $L_{\omega_1^{CK}}$ in particular; Sacks' book is as usual a good source on the topic.