Shell – Check if content of file has length X and contains only specific characters

One way is to use cut:

 command | cut -c1-8

This will give you the first 8 characters of each line of output. Since cut is part of POSIX, it is likely to be on most Unices.

You can make a here document and use cat:

$ cat <<EOT >output
$a
EOT
$ wc -l output
40000

I don't know exactly how powerful your Busybox's ash is (it's configurable), but that should work anywhere, even with no builtins at all. It expands the variable value into a quasi-file, which is then given to cat, rather than putting the value itself into the command line.

Since you seem to have literal "\n" in your string that you want to get rid of, you can also use string replacement during parameter subsitution. This can be configured out of Busybox's ash, in which case you'll have to go to sed. If you do have this (non-POSIX) extension, you can use ${var//\\\n/$NEWLINE}:

$ busybox ash
$ foo="x\ny\nz"
$ echo "$foo"
x\ny\nz
$ NEWLINE="
"
$ echo "${foo//\\\n/$NEWLINE}"
x
y
z

This is the same syntax as Bash; ${var/pattern/replacement}, where pattern is global if it starts with /. It seems your ash has a lot configured out of it, so this might not be there at all. In that case you're probably going to have to go with sed, although I guess judicious use of IFS and read would let you work around that and reconstruct a correct string.