Can the Unix list command 'ls' output numerical chmod permissions?

it almost can ..

 ls -l | awk '{k=0;for(i=0;i<=8;i++)k+=((substr($1,i+2,1)~/[rwx]/) \
             *2^(8-i));if(k)printf("%0o ",k);print}'

Closest I can think of (keeping it simple enough) is stat, assuming you know which files you're looking for. If you don't, * can find most of them:

/usr/bin$ stat -c '%a %n' *
755 [
755 a2p
755 a2ps
755 aclocal
...

It handles sticky, suid and company out of the box:

$ stat -c '%a %n' /tmp /usr/bin/sudo
1777 /tmp
4755 /usr/bin/sudo

you can just use GNU find.

find . -printf "%m:%f\n"