Does $\#W$[1]-hardness imply approximation hardness?
$\mathrm{W}[1]$-hardness implies that a problem has no eptas unless (at least) $\mathrm{W}[1] = \mathrm{FPT}$ (having an eptas implies parameterized tractability for the standard solution size parameterization), but there are problems with a ptas that are $\mathrm{W}[1]$-hard (i.e. not $\mathrm{APX}$-hard unless $\mathrm{APX} = \mathrm{PTAS}$).

Transferring this to $\#\mathrm{W}[1]$-hardness, you can at least say that a $\#\mathrm{W}[1]$-hard problem still has no eptas, but I don't think a stronger statement can be made a priori. Turning to speculation, I suspect it's also not true that $\mathrm{APX}$-hardness immediately follows from $\#\mathrm{W}[1]$-hardness in general, though perhaps something may be said for higher classes, such as the counting version of $\mathrm{para}\text{-}\mathrm{NP}$.