The $[[5,1,3]]$ code is a perfect code basically meaning that the weight-0 and weight-1 error spaces completely fill out the $32$-dimensional Hilbert space.
On the other hand, the $[[7,1,3]]$ Steane code is not perfect. There are $6$ stabilizer generators amounting to $2^6 = 64$ syndromes and yet the weight 0/1 errors only account for $22$ of these, leaving $42$ unused syndromes!
In practice it seems prudent to assign some of the weight-2 errors to these $42$ unused syndromes. However, there are $3^2 \binom{7}{2}=189$ weight-2 errors so we certainly cannot have a syndrome for all of these errors (which makes sense because in that case the code would have distance $d=5$ instead). This means we must make a choice for which weight-2 errors we wish to correct. I have never seen this done but I am sure that someone has done it somewhere (does anyone have a reference)?
Is there some choice that results in a better logical error rate than some other choice? Of course it depends on the error model, for example if a $Z$ error is more likely than an $X$ or $Y$ error then you probably want a syndrome for each of the weight-2 $Z$-type errors and maybe even some weight-3 $Z$-type errors. But in the special case of a depolarizing channel (where $X$, $Y$, and $Z$ occur with the same probability) I would assume that the logical error rate is invariant under this choice. Am I correct?
Edit (2-29-2024): As discussed in the answer by @ChrisD, there seems to be a canonical choice for weight-2 syndromes for the Steane code owing to the fact that it is a CSS code. However, does this choice lead to the best probability of logical error (or more simply, does this choice lead to the best pseudo-threshold) amongst all possible choices of weight-2 syndromes?