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    $\begingroup$ I don't have a rigorous answer, but it appears that it is linked to the fact f is not defined. If you define f (e.g. f[l___]:=1 the result of NumericQ@f[{1},2,3] become True $\endgroup$ Commented May 14, 2020 at 13:21
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    $\begingroup$ @youyou Isn't it because f evaluates to 1 first? NumericQ doesn't have any Hold* attributes. $\endgroup$ Commented May 14, 2020 at 15:27