Remove leading string in bash

The commands you passed to sed mean: if a line matches the regex, delete it. That's not what you want.

echo "$TEST" | sed 's/rev0*//'

This means: on each line, remove rev followed by any number of zeroes.

Also, you don't need sed for such a simple thing. Just use bash and its parameter expansion:

shopt -s extglob         # Turn on extended globbing.
echo "${TEST##rev*(0)}"  # Remove everything from the beginning up to `rev`
                         # followed by the maximal number of zeroes.

POSIXly:

test='rev00000010'
number=${test#"${test%%[1-9]*}"}

Would remove every thing to the left of the left-most non-zero digit.

Bournely/universally:

number=`expr "x$test" : 'xrev0*\(.*\)'`

If you already have the variable TEST just print it with all letter removed

printf "%.0f\n" ${TEST//[a-z]/}

or

printf "%g\n" ${TEST//[a-z]/}

Do not use %d or echo-command becouse numbers with leading 0 is understood as octal