It would be interesting to see what's in $_ at that moment. From perldoc -f printf (emphasis mine):

printf FILEHANDLE FORMAT, LIST
printf FORMAT, LIST

Equivalent to "print FILEHANDLE sprintf(FORMAT, LIST)", except that "$\" (the output record separator) is not appended. The first argument of the list will be interpreted as the "printf" format. See "sprintf" for an explanation of the format argument. If "use locale" is in effect, the character used for the decimal point in formatted real numbers is affected by the LC_NUMERIC locale. See perllocale.

Don't fall into the trap of using a "printf" when a simple "print" would do. The "print" is more efficient and less error prone.

Try Use

        print STDOUT $_;
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Though I can't conceive a way in which $_ could be seen as a format that touches some read-only value while interpolating an empty list, it could be that's just what's happening here.

update: Try

perl -le '$_ = "%n"; printf STDOUT $_'
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hmm, always late today...

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}