Use variables in here-string

That's not the here string, it's ANSI-C quoting:

Words of the form $'string' are treated specially. ... The expanded result is single-quoted, as if the dollar sign had not been present.

So what you've got is a single-quoted string to the right of <<<. That string gets taken as the here string, with no further processing.

There's no need to use only a single set of quotes around the whole word, however. You can use several quoted parts (or unquoted single words) joined together:

script <<< "$var 2"$'\n'"3 4"$'\n'"5 6"$'\n'q

will do what you wanted.

Alternatively, you could backslash-escape the spaces, rather than quoting "1 2".

You could also use echo -e with a regular pipe:

echo -e "$var 2\n3 4\n5 6\nq" | script

or printf:

printf '%s 2\n3 4\nq' "$var" | script

-e is necessary to enable escape processing in echo's arguments. printf does those by default, but it has its own interpolation system (printf "$var 2\n3 4\nq" will also work, but is problematic if $var might contain escape characters).

The $'...' ANSI-C quoting results in a "single-quoted" string.
Inside a single quoted string variables are not expanded:

$ var=13

$ echo 'test $var'
test $var

You need to take the variable out of such structure. The easiest way IMO is:

$  printf -v val '%s 2\n3 4\n5 6\nq' "$var"
13 2
3 4
5 6
q

The format to printf is very similar to what you had before, and the %s is replaced by the value of the variable. To get the result in a variable, we can use the capacity of bash printf of setting a variable:

$ printf -v val '%s 2\n3 4\n5 6\nq' "$var"
$ script <<< "$val"

All in one line:

$ printf -v val '%s 2\n3 4\n5 6\nq' "$var"; script <<< "$val"