Arithmetic statement which is independent, and whose independence is independent, and so on?

If $p$ is independent of ZFC, and ZFC is sound, then $I(p)$ is independent of ZFC.

Indeed, because ZFC is sound, it can't disprove $I(p)$ because $I(p)$ is true.

However, $I(p)$ implies consistency of ZFC, which isn't provable in ZFC by Godel's theorem (using soundness again), so $I(p)$ is not provable.

Thus $I(p)$ is independent.

If we fix things to avoid Will Sawin's observation, then the answer is yes under any reasonable interpretation I can think of.

For example, consider the following: let $J(p)$ be the sentence "If $\mathsf{ZFC}$ is consistent then $p$ is independent over $\mathsf{ZFC}$." We can indeed prove in $\mathsf{ZFC}$ many instances of $J$ (e.g. that $J$ holds of the Godel-Rosser sentence for $\mathsf{ZFC}$), so this is nontrivial.

However, note that if we assume that $\mathsf{ZFC}$ is sound (this is overkill but let's do it anyways), then we have $\mathsf{ZFC}\vdash J^n(p)$ for some finite $n$ only if $p$ really is independent of $\mathsf{ZFC}$. Since the set of sentences which are independent over $\mathsf{ZFC}$ is not c.e., there must be many $p$s such that $\mathsf{ZFC}$ doesn't prove any of the $J^n(p)$s.

• The non-c.e.-ness claim above is an instance of a more general fact: if $T$ is any theory to which Godel's first incompleteness theorem applies, then the set of $T$-independent sentences is not c.e. This is because if it were, then the set of $T$-theorems would be both c.e. (because it obviously is) and co-c.e. (because every non-$T$-theorem is either $T$-disprovable, which is obviously a c.e. condition, or is $T$-independent which is a c.e. condition by assumption). But that would make the set of $T$-theorems computable, which can't be the case (e.g. if it happened we'd be able to whip up a computable complete extension of $T$).

Basically, any reasonable iterated independence property can't fully capture independence.