my $analysis_file = shift;

open INFILE_1, '<', $analysis_file
    or die "Can't open '$analysis_file': $!\n";

while(<INFILE_1>) {
    my $ALL_FT_FILE = $_; # or some part of $_ after munging

    open INFILE_2, '<', $ALL_FT_FILE
        or die "Can't open '$ALL_FT_FILE': $!\n";

    while(<INFILE_2>) {
        ...  # is the outer $_ clobbered here?
    }
}
[download]

That's considerably shorter, and more to the point, isn't it?

Answer: No, each $_ is localized to its own while() block. But you can't access the contents of the outer $_ from the inner block, unless you assign that to another variable.

<update>

*Sigh*. As jdporter pointed out, yes, $_ isn't localized in a while loop:

open O, '>', 'foo';
print O "bar\n";
close O;
open O, '>', 'bar';
print O "$_\n" for qw(foo quux wrdlbrmfd);
close O;

open I, '<', 'foo';
while (<I>) {
    chop;
    open II, '<', $_ if -f $_;
    while (<II>) {
        print;
    }
    print "\$_ = '$_'\n";
}
__END__
foo
quux
wrdlbrmfd
$_ = ''
[download]

As you see at the last line of the ouput above, $_ is clobbered.

</update>

Some nits:

• three argument open is generally preferred (there are edge cases where 2-argument open makes sense)
• the construct "${ALL_FT_FILE}"is only used to disambiguate an interpolated variable from an adjacent literal, as in "${foo}_bar"; interpolating a single variable in a double quoted string is bad practice, unless it is necessary - in your code it is not
• code inside a block is better readable if it's indented relative to its enclosing brackets

</pedantic> :)

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