'&' is bit-wise boolean and. The expression ($i & 2**$j), is testing the binary value of $i and returns true if the $jth bit is set to one.

This may be clarified by this slightly modified version of your example code:

[11:04:18.32] P:\test>perl
my $n = 4;
for my $i (1..2**$n-1) {
    printf "%2d (0b%04b) has bits [ ", $i, $i;
    for my $j (0..$n-1) {
        if ($i & 2**$j) {
            print $j, " ";
        } # end-if
    } # end-for
    print "] set\n";
} # end-for
^Z
 1 (0b0001) has bits [ 0 ] set
 2 (0b0010) has bits [ 1 ] set
 3 (0b0011) has bits [ 0 1 ] set
 4 (0b0100) has bits [ 2 ] set
 5 (0b0101) has bits [ 0 2 ] set
 6 (0b0110) has bits [ 1 2 ] set
 7 (0b0111) has bits [ 0 1 2 ] set
 8 (0b1000) has bits [ 3 ] set
 9 (0b1001) has bits [ 0 3 ] set
10 (0b1010) has bits [ 1 3 ] set
11 (0b1011) has bits [ 0 1 3 ] set
12 (0b1100) has bits [ 2 3 ] set
13 (0b1101) has bits [ 0 2 3 ] set
14 (0b1110) has bits [ 1 2 3 ] set
15 (0b1111) has bits [ 0 1 2 3 ] set
[download]

The first column is the value of $i in decimal. The second is the value of $i as binary. The numbers in []s are the positions (numbered right to left, from zero) of the bits that are set in the current value of $i.