I share Laurent_R's confusion about the structure of your data. Here are some general thoughts.

Does this make hash into hashes ...

Why not see for yourself?

c:\@Work\Perl\monks>perl -wMstrict -le
"use Data::Dumper;
 ;;
 $_ = '-1.23,4...5,6,foo,bar,baz,boff';
 ;;
 my %hash;
 ;;
 if (/^\s*(-\d+\.*\d*),(\d+\.*\d*),(\d+\.*\d*),(\S+),(\S+),(\S+),(\S+)
+$/) {
   my $data1 = $1;
   my $data2 = $2;
   my $data3 = $3;
   my $data4 = $4;
   my $data5 = $5;
   my $data6 = $6;
   ;;
   $hash{data2}{data1}{data3} = $data3;
   $hash{data2}{data1}{data4} = $data4;
   $hash{data2}{data1}{data5} = $data5;
   $hash{data2}{data1}{data6} = $data6;
   }
 ;;
 print Dumper \%hash;
"
$VAR1 = {
          'data2' => {
                       'data1' => {
                                    'data6' => 'baz',
                                    'data4' => 'foo',
                                    'data3' => '6',
                                    'data5' => 'bar'
                                  }
                     }
        };
[download]

(See the Data::Dumper module, which is core, for more info.) Is this the data structure you expected? Did you perhaps mean to write
    $hash{$data2}{$data1}{$data3} = $data3;
instead of
    $hash{data2}{data1}{data3} = $data3;
($data1 $data2 are unused otherwise)?

Are you aware that the regex
    /^\s*(-\d+\.*\d*),(\d+\.*\d*),(\d+\.*\d*),(\S+),(\S+),(\S+),(\S+)$/
has seven capture groups, but you only use six ($data6)? Are you aware that the regex expression  \.* matches zero or more  . (dot) characters (i.e., it matches '.....')?