What you are observing is exactly what dynamic scoping is about: Code called from where the variable is defined can see it, even though it's not in the same lexical scope.

This can be demonstrated without recursion:

use 5.010;
use strict;
use warnings;

sub sayit {
    say $1 // 'undef';
}

do {
    '42' =~ /(\d+)/ and sayit();
};
sayit();
__END__
42
undef
[download]

Subroutine sayit reads $1 outside of the lexical scope of the block where $1 is set. But since it's in the dynamic scope, it can still see the value.

The sayit call outside the block prints undef\n, which demonstrates that $1 isn't merely a global variable.

As to your actual question:

However, $1, set to '1' by the last successful match at the lowest-but-one level of recursion, is propagated upward unchanged through several levels of subroutine 'blocks'

I see no evidence for that. After $1 is set to '1', exactly one more recursive call happens, and there it is printed out. Then the recursion ends, and you don't print $1 anymore.

Update: after experimenting a bit, I can provide evidence of the phenomen you mentioned:

sub R {
   printf qq{before: \$_ is '$_'};
   printf qq{  \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined';
   s/(\d+)// ? $1 + R() : 0;
   printf qq{after: \$_ is '$_'};
   printf qq{ \$1 is %s \n}, defined($1) ? qq{'$1'} : 'undefined';
}

$_ = 'x55x666x7777x1x';
R();
__END__
before: $_ is 'x55x666x7777x1x'  $1 is undefined 
before: $_ is 'xx666x7777x1x'  $1 is '55' 
before: $_ is 'xxx7777x1x'  $1 is '666' 
before: $_ is 'xxxx1x'  $1 is '7777' 
before: $_ is 'xxxxx'  $1 is '1' 
after: $_ is 'xxxxx' $1 is '1' 
after: $_ is 'xxxxx' $1 is '1' 
after: $_ is 'xxxxx' $1 is '1' 
after: $_ is 'xxxxx' $1 is '1' 
after: $_ is 'xxxxx' $1 is '1' 
[download]

This is because there is only one variable $1. It is dynamically scoped, so once it is set in an inner scope, an outer scope sees the modification too.