What is the minimum required depth of reductions for NP-hardness of SAT? As everyone knows, 
SAT is complete for $\mathsf{NP}$ w.r.t. polynomial-time many-one reductions. 
It is still complete w.r.t. $\mathsf{AC^0}$ many-one reductions.

My questions is what is the minimum required depth for the reductions? 
More formally, 


  What is the least $d$ such that 
  SAT is $\mathsf{NP}$-hard w.r.t $\mathsf{AC^0_d}$ many-one reductions?


It seems to me that $\mathsf{AC^0_2}$ should be sufficient? Does anyone know a reference?
Reposting my comment:

From a quick look, it seems like your question should be answered by "Manindra Agrawal, Eric Allender, Steven Rudich, Reductions in Circuit Complexity: An Isomorphism Theorem and a Gap Theorem, JCSS 57: 127-143, 1999." They say "we prove that all sets complete for NP under AC0 reductions are complete under reductions that are computable via depth two AC0 circuits." But I may be missing something.