Is there any interesting consequence of $\mathrm{DLogTime}$-uniform ${\mathrm{Mod}_6}^0=\mathrm{NP}$
Even a weaker claim that $\mathrm{DLogTime}$-$\mathrm{uniform}\:\mathrm{AC_2^0}=\mathrm{L}$ readily implies $\mathrm{NP}= \mathrm{PSPACE}$

First, $\mathrm{NP} = \mathrm{EXP}$-$\mathrm{AC_2^0}$ by using a similar argument in this answer. Here, I mean constant-depth $\mathrm{AC}^0$ circuit of depth at most $2$. And, $\mathrm{PSPACE} = \mathrm{EXP}$-$\mathrm{L}$, if you are willing to abuse the notation.

Second, since the mentioned above answer does not describe in details the uniformity of the first-order reduction. I move on to prove the $\mathrm{\mathrm{DLogTime}}$-$\mathrm{uniformity}$ of the standard Cook proof.

That is we need to establish a $\mathrm{DLogTime}$ constructor that constructs a constant-depth circuit, for each input length. And the constructed constant-depth circuits then receives arbitrary input (of given length) and always output correctly.

Indeed, Cook proves completeness of $\mathrm{SAT}$ by using a big tableau.

From the second row of the tableau on, every cell can be computed by a constant-size formula of "nearby" cells of previous row. So, these can be printed as constant output node of the constructed constant-depth circuit. This can be called hard-coded. So, almost all of the constructed constant-depth circuit are merely constant output nodes.

We only need to deal with the first row. This part of the constructed circuit needs to take the actual input bits as its own designated input gates. So, it requires some more indexing work of the constructor. It is worthy mentioning that these actual input bits does not contribute to the variable of the formula output by our constructed circuit (the circuit is a constant-depth reduction, not a decider). This very first part of the constructed circuit can be constructed by the $\mathrm{DLogTime}$ constructor easily. What really makes it into variables of the formulae that will be reduced to by our constructed circuit are the nondeterministic guess bits (also called the witness if you prefer the verifier-style definition of $\mathrm{NP}$). This witness part can also be printed out as constant output gates of the constructed circuit.

Back to your question, we can easily adapt the above argument to deduce that the strange consequence of $\mathrm{DLogTime}$-$\mathrm{uniform}$ $\mathrm{Mod}_6^0 = \mathrm{NP}$ is that $\mathrm{Mod}_6PH = NEXP$. If someone can prove that $\mathrm{Mod}_6P$ is low for itself like $\oplus P$ then the consequence can be strengthened to $\mathrm{Mod}_6P = NEXP$