The Feferman-Schutte analysis is completely wrong. The problem is that "well-ordered" is not a single predicative concept because predicativists lack strong comprehension axioms. If you can't form the set $\{x \in X: \neg\phi(x)\}$ then you can't go from "every subset of $X$ has a smallest element" to "if $(\forall x < a)\phi(x) \to \phi(a)$ for any $a \in X$,then $(\forall x \in X)\phi(x)$".
This wrecks their whole argument because at each stage they use proof trees of height $\gamma_n$ to prove that $\gamma_{n+1}$ is well-ordered in the weak sense and then claim that this justifies acceptance of proof trees of height $\gamma_{n+1}$, which would require induction up to $\gamma_{n+1}$ for all sentences of second order arithmetic.
The argument is absurd in the other direction as well --- if the predicativist can grasp the jump from $\gamma_n$ to $\gamma_{n+1}$ for each $n$, why can't he grasp the general principle that for all $n$, induction up to $\gamma_n$ implies induction up to $\gamma_{n+1}$, and infer induction up to $\Gamma_0$? No cogent answer to this question was ever given. Believe me, I dug through the literature pretty hard.
I discuss this in more detail than you could possibly want in my paper Predicativity beyond $\Gamma_0$.
The axioms of first-order arithmetic include the induction schema, which says that, for every formula $A(x)$ with free variable $x$, the conjunction of $A(0)$ and $\forall x\,(A(x)\rightarrow A(x+1))$ implies $\forall x\,A(x)$. This is, of course, a special case of the well-known and basic induction property of the natural numbers that says the same thing for any property $A(x)$ whatsoever, whether or not it's defined by a first-order formula. For anyone who (1) understands the natural numbers well enough to grasp the general induction principle and (2) believes that (first-order) quantifiers over the natural numbers are meaningful so that first-order formulas $A(x)$ really define properties, it is clear that the natural number system satisfies all of the first-order Peano axioms, and therefore those axioms are consistent.
A difficulty arises if one adopts a very strong constructivist or finitist viewpoint, doubting item (2) above, i.e., questioning the meaning of first-order quantifiers $\forall z$ and $\exists z$ when $z$ ranges over an infinite set (like $\mathbb N$) so that one can't actually check each individual $z$. From such a viewpoint, the formulas $A(x)$ occurring in the induction schema are gibberish (or close to gibberish, or at least not clear enough to be used in mathematical reasoning), and then the proposed consistency proof collapses.
The chief virtue of Gentzen's consistency proof is that it essentially avoids any explicit quantification over infinite sets. It can be formulated in terms of very basic, explicit, computational constructions (technically, in terms of primitive recursive functions and relations). There is, however, a cost for this virtue, namely that one needs an induction principle not just for the usual well-ordering of the natural numbers but for the considerably longer well-ordering $\varepsilon_0$.
Thus, Gentzen uses a much longer well-ordering, but his induction principle is only about primitive recursive properties, not about arbitrary first-order definable properties. There is a trade-off: Length of well-ordering versus quantification.
I believe the trade-off can be made rather precise, but I don't remember the details. Recall that $\varepsilon_0$ is the limit of the sequence of iterated exponentials $\omega(0)=\omega$ and $\omega(n+1)=\omega^{\omega(n)}$. If we weaken PA by limiting the induction principle to formulas $A(x)$ that can be defined with a fixed number $n$ of quantifiers, then the consistency of this weakened theory can be proved using primitive recursive induction up to $\omega(n)$, as proved by Carnielli and Rathjen in "Hydrae and subsystems of arithmetic". In other words, the trade-off is that an additional quantifier in the induction formulas costs an additional exponential in the ordinal.
Best Answer
I believe the answer is "no": certainly the proof-theoretic ordinal (the optimal object taking the role of "$\epsilon_0$" in Gentzen's proof) is totally unknown, and my understanding is that there is no non-trivial upper bound on it, either. See also Proof-Theoretic Ordinal of ZFC or Consistent ZFC Extensions?; in particular, my understanding is that we only know the actual proof-theoretic ordinals of theories up to (something around) $\Pi^1_2$-comprehension, and I suspect we don't even have upper bounds for proof-theoretic ordinals as high as $\Pi^1_3$-comprehension.
That said, I also believe this is the only obstacle - that is, if we had the proof-theoretic ordinal $\alpha$ in hand, then we would have a proof of "Analysis is consistent" from "$T$+'induction along $\alpha$'", where $T$ is some reasonable weak base theory and "induction along $\alpha$" is properly formulated. So in some sense, the only real difference between analysis and arithmetic is in the complexity of finding the proof-theoretic ordinal.
. . . however, the use of the word "only" there can be extremely misleading: finding proof-theoretic ordinals of stronger and stronger theories requires increasingly deep ideas, and not just the continued turning of a well-understood crank. So on the one hand, while it's valuable to localize all the difficulty around a single object, it's also true that this is a vastly complex object, and that saying "all we need to do is understand the proof-theoretic ordinal" is less meaningful than it might sound.