$\newcommand\Con{\text{Con}}
\newcommand\Dec{\text{Dec}}$


Let $F$ be the formal system in which the proofs are to be carried
out, when it comes to your formal assertions of the form
$\Dec(\varphi)$. So we assume that $F$ is described by some
computable axiomatization. For example, perhaps $F$ is simply the
usual first-order PA axioms. Let me assume that $F$ is true in the
standard model $\mathbb{N}$, which is probably a case that you care most about. (But actually, I believe it is
sufficient in this argument to assume iterated consistency assertions about $F$.)

Let $A=\Con(F)$. I claim that this statement is
$\infty$-undecidable with respect to $F$.

To see this, argue as follows. By the incompleteness theorem,
since $\Con(F)$ is true, we know that $F$ does not prove $A$, and
since $F$ and $A$ are both true, it also follows that $F$ does not
prove $\neg A$. So $\neg\Dec(A)$ is true (that is, in the standard
model $\mathbb{N}$). But $F$ by itself cannot prove $\neg\Dec(A)$,
since $F$ proves $\neg\Dec(X)\to \Con(F)$, as an inconsistent theory has no undecidable statements, and so if it did it
would violate the incompleteness theorem. Note also that $F$
cannot prove $\Dec(A)$ either, since $\neg\Dec(A)$ is true. Thus,
$\neg\Dec(\Dec(A))$ is true. But $F$ cannot prove this, since then
again it would prove $\Con(F)$, violating incompleteness, and it
also cannot prove $\Dec(\Dec(A))$, since $\neg\Dec(\Dec(A))$ is
true. So $\neg\Dec(\Dec(\Dec(A)))$ is true. And so on.

This reasoning shows that $\Dec^n(A)$ will be false for every $n$,
and so $A=\Con(F)$ is $\infty$-undecidable, assuming that $F$ is
consistent and true in the standard model.

It seems likely to me that the content of what it was about "true
in the standard model" that the argument used should be covered by
the assumption merely that $F+\Con^n(F)$ holds for all $n$. But I
shall leave this to the proof-theoretic experts, who I hope will
shed light on things.