Firstly, you omitted the other contender:

...
sub lookbehind {
    my $x = shift;
    split '(?:(?<=[^#]####)|(?<=[^#]##)|(?<!#))[@]', $x;
}

C:\test>junk8
Subroutine Benchmark::mytime redefined at c:/Perl/lib/Benchmark.pm lin
+e 459.
Benchmark:
timing 10000 iterations of
 lookbehind, rollahead, unroll
...
lookbehind: 2.34183 wallclock secs ( 2.19 usr +  0.00 sys =  2.19 CPU)
+ @ 4570.38/s (n=10000)
 rollahead: 3.46343 wallclock secs ( 3.23 usr +  0.00 sys =  3.23 CPU)
+ @ 3092.15/s (n=10000)
    unroll: 3.51775 wallclock secs ( 3.33 usr +  0.02 sys =  3.34 CPU)
+ @ 2990.43/s (n=10000)

             Rate     unroll  rollahead lookbehind
unroll     2990/s         --        -3%       -35%
rollahead  3092/s         3%         --       -32%
lookbehind 4570/s        53%        48%         --
[download]

Secondly, you didn't check your results. unroll and rollahead produce different results:

          rollahead => @# #@## ##@  ##@## #@# # @# #@## ##@@# #@## ##@
+@# #@ ...
             unroll => @# #@## ##@  ##@## #@#  @# #@## ##@@# #@## ##@@
+# #@# ...
[download]

And both sets are wrong! Any output that has an even number of #s preceding a @ (as with the 3rd, 4th etc. outputs above) is wrong. If you look back at the op, you'll see that any @s that persist into the results set are preceded by an odd number of ##s.

The results set from the fastest algorithm:

          lookbehind => #@## ###@#### #####@  #####@#### ###@## # #@##
+ ###@#### ...
[download]