-n wraps your code in while (<>) { ... }* (cf perldoc perlrun). Thus, your one-liner is equivalent to:
perl -e '
while(<>) {
%ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }
}
'
[Line breaks added for readability. They can be removed if you so desire.]
See the double while (<>) { ... }? That's your issue: the first while (the one added by -n) reads a line, then the second while (the one you wrote) reads a second line, does your s/// (on the second line), and prints this second line updated. Thus, you need a blank line before the actual line you want to process.
To fix the issue, either remove the inner while(<>), or remove the -n flag. For instance:
perl -e '
%ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m");
while (<>) { s/(<([BINORSU])>)/\e$ES{$2}/g; print; }
'
Or,
perl -ne '
BEGIN { %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m") };
s/(<([BINORSU])>)/\e$ES{$2}/g; print;
'
Note that instead of using -n and print, you can use -p, which is the same as -n with an extra print** at the end:
perl -pe '
BEGIN { %ES=("B","[1m","I","[3m","N","[m","O","[9m","R","[7m","U","[4m") };
s/(<([BINORSU])>)/\e$ES{$2}/g;
'
* For completness, note that -n adds the label LINE before the while loop (LINE: while(<>) { ... }), although that doesn't matter in your case.
** The print added by -p is actually in a continue block after the while, although, once again, this doesn't matter in your case.