Examples of provably${}^n$ unprovable statements

[This is a partial answer, but I decided to post it anyway since no one else has answered for a long time now. Very interesting question by the way!]

Note that if $T = PA + \neg Con(PA)$, then $T \vdash \neg Con(T)$ and hence $T_2$ is inconsistent and proves everything, so your hierarchy collapses. So it's not enough just to assume that $T$ contains $PA$. I think you need the much stronger condition of assuming that $T$ is arithmetically sound, namely that every arithmetical statement that $T$ proves is true for $\mathbb{N}$. This would happen if for example $T$ has an $ω$-model, namely that there is a model of $T$ where the arithmetical part is $\def\nn{\mathbb{N}}$$\nn$. If the language of $T$ is the language of arithmetic (with no extra symbols), then these two notions are equivalent and just mean that $\nn$ satisfies $T$, but in general arithmetic soundness is weaker.

So we shall from now assume that $T$ has decidable proof validity and is arithmetically sound. Note that by induction $T_n$ is arithmetically sound for every $n \in \nn$.

Let $D(n)$ denote the class of sentences at level $n$ in your hierarchy, and let $D(m,n)$ denote your class $m,n$. And let $D^*(k,l) = \bigcup \{ D(m,n) : k \le m \le ω \land l \le n \le ω \}$.

Firstly, your statement that independent sentences are in $D(2,2)$ is not correct, because all you know is that it is in $D^*(2,2)$, since the very definition of independence is equivalent to being in $D^*(2,2)$. Note that our entire discussion is in some meta-system, usually taken to be ZFC. The question of whether or not a sentence over $T$ is independent over $T$ has a fixed truth-value in the meta-system. Even if we cannot prove some sentence independent, it does not change which class it falls in.

Some sentence over $T$ is in $D^*(2,3)$.

Claim: $Con(T_2) \notin D(1)$. Proof: If $T_1 \vdash Con(T_2)$ then $T_1 \vdash Con(T_1)$, which gives a contradiction.

Claim: $( \neg Con(T_2) ) \notin D(1)$. Proof: If $T_1 \vdash \neg Con(T_2)$, then $T_1 \vdash ( T_1+Con(T_1) \vdash \bot )$, and hence $T_1 \vdash ( T_1 \vdash \neg Con(T_1) )$, but by arithmetical soundness this implies $T_1 \vdash \neg Con(T_1)$, which gives a contradiction.

Claim: $( \neg Con(T_2) ) \notin D(2)$. Proof: If $T_2 \vdash ( T_1 \nvdash \neg Con(T_2) )$, then $T_2 \vdash Con( T_1 + Con(T_2) )$, and hence $T_2 \vdash Con(T_2)$, which gives a contradiction.

Claim: $Con(T_2) \in D(2)$. Proof: $T_2 \vdash ( T_1 \nvdash Con(T_1) )$, and hence $T_2 \vdash ( T_1 \nvdash Con(T_2) )$.

Conjecture: $( \neg Con(T_2) ) \in D(3)$.

Therefore $Con(T_2) \in D^*(2,3)$, and I think it is in $D(2,3)$.

I also think that there is some sentence $φ$ over $T$ that is in $D^*(ω,ω)$, but don't see why.