Weakest reduction for 3-$\mathrm{SAT}$
Having read all your answers to my questions, I come to the ultimately weakest reduction for 3-$\mathrm{SAT}$

The weakest reduction is $\mathrm{NC}_0^0$. This reduction does not use even $\mathrm{NOT}$  gate. Absolutely, almost nothing at all.

Consider the very first part of the very first row in your answer here. As you mentioned there, this only needs some indexing work of the $\mathrm{DLogTime}$-constructor. It turns out that all that is needed to be done is to put $\lnot$ symbol to the soon-to-be-reduced-to formula by copying the actual input bits themselves.

In particular, consider the 10th cell $c_{10}$ of the first row, this cell is equal to $x_{10}$ i.e. the 10th bit of the actual input.

Depending on whether $x_{10}$ is 0 or 1, the corresponding clause will be:

$$ \lnot c_{10} \lor \lnot c_{10} \lor \lnot c_{10} $$

or

$$ c_{10} \lor c_{10} \lor c_{10} $$

All the $c_{10}$'s are hard-coded similarly as the rest of the constructed constant-depth circuit. The $\lnot$ or nothing (actually a dummy symbol in the soon-to-be-reduced-to formula) is copied from the actual input bit $x_{10}$.

So, the weakest reduction is $\mathrm{NC}_0^0$. It copies the input bits to some output gates of the constructed 0-depth circuit. All the other output gates of the constructed 0-depth circuit are constantly hard-coded.