Not directly (maybe using PadWalker or some such), unless you make it available to the outside, e.g. by returning it as an (optional) second value:

#!/usr/bin/perl -l
use strict;
use warnings;

my $code_ref = sub
{
    my $array_ref;
    my $nested_code_ref = sub
                          {
                               my @array = ("one","two","three");
                               $array_ref = \@array;
                          };
    $nested_code_ref->();
    return wantarray ? ($array_ref, $nested_code_ref) : $array_ref;
};

print $code_ref->()->[0];
print $code_ref->()->[1];

print( ( $code_ref->() )[1]->()[2] );

__END__
$ ./837730.pl
one
two
three
[download]

We might be able to give a more useful answer, if you could explain your intention — from the question it's not really clear whether you want achieve or avoid that $nested_code_ref be accessible from the outside...

Is it possible to call the anonymous block referenced by $nested_code_ref outside the anonymous block $code_ref.

No. Unless you store it in a variable where it's accessible from the outside. (Or use something evil like PadWalker, but that's considered cheating).