This took me a little while to figure out how to solve, but it's really very easy. While there are still items to order, look for items that come before only items that have already been ordered. Insert them at start (or at end and reverse, like I did). The algorithm is at worst case about (N^2) / 2, efficient enough for the task given here.

use strict;
use warnings;

my %before = (
    a => ['c'],
    b => ['d','e'],
    c => ['l'],
    d => ['e','a'],
    e => ['c'],
    f => ['g','d'],
    g => ['c'],
    h => ['g','i'],
    i => [],
    j => ['c'],
    k => [],
    l => [],
    n => ['c'],
    o => [],
);

my ($k, $m, %p, @order);
while (scalar keys %before) {
    for $k (keys %before) {
        $m = 1;
        for (@{$before{$k}}) {
            if (!exists $p{$_}) {
                $m = 0; last;
            }
        }
        if ($m) {
            push(@order, $k);
            $p{$k} = ();
            delete($before{$k});
        }
    }
}
@order = reverse @order;
print @order;
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