Well, I would characterize the reason as having more to do with the fact that passing $a = 2 as an argument causes an alias to $a to be put into @_. Had a copy of $a's value been placed into @_ instead, then subsequently changing $a to 3 would not impact that copy.

print $a = 1, 0+$a, $a = 2, 0+$a, $a = 3, 0+$a;
# prints "313233"
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The reason that $a = 2 passes an alias is because it is an "lvalue" expression, which just means that you can do previously mentioned things like ( $a = 2 ) = 3. And the reason that lvalue expressions cause aliases to be passed is two-fold: So you can modify arguments via code like $_[0] = 3 in your sub and because it is more efficient than creating a copy.

This is because the entire expression is evaluated before being passed as arguments to print

No, the arguments are placed on the stack as they are evaluated.

1. $a = 2 is evaluated and the result ($a) is placed on the stack.
2. ' ' is evaluated and the result is placed on the stack.
3. $a = 3 is evaluated and the result ($a) is placed on the stack.
4. "\n" is evaluated and the result is placed on the stack.
5. print is evaluated.

@_ isn't even constructed here since print is an operator, not a sub.

The reason you get the observed behaviour is that $a itself (not a copy of it) is placed on the stack.