What's wrong

for filename in $(ls -1 ./*); do

Ouch! As a general rule (with very few exceptions), do not use ls in scripts. What you wrote is almost equivalent to for filename in ./*; do, except that if there are any non-printable characters, whitespace characters, or \[?* in the file names, they will be mangled if you use ls. You don't need the ./ (except to ensure that the file name doesn't start with a -) but it doesn't hurt.

[[ $filename =~ _[0-9]\. ]]

could be written in a slightly simpler and more portable way: [[ $filename = *_[0-9].* ]]. And since you can use a simple glob pattern, you might as well not have iterated over all file names: for filename in *_[0-9].*; do. But there's a better way to express your script, to take advantage of BASH_REMATCH; see below.

echo $filename

Always put double quotes around variable and command substitutions. Exception: when you understand why you need to leave the double quotes off and why it's safe to do so. When the shell sees a variable substitution ($foo or ${foo}) or a command substitution (`foo` or $(foo)) outside double quotes, the result of the substitution undergoes word splitting and globbing (filename generation). That was one of the problems with $(ls -1 ./*) earlier. This should be echo "$filename".

In fact, it would be better printf "%s" "$filename", because echo itself performs expansions. In bash, unless you've set non-default options to enable backslash expansion, the only problem is that a few arguments beginning with a - look like option, and in this specific case the filename will begin with ./.

There is an edge case where your sed call won't work: if you have a file name that ends with a newline character. This doesn't happen in practice unless someone has made a mistake (like a rogue script or a bad copy-paste) or is deliberately trying to trick your script — so watch out in security contexts.

Incidentally, this is one of the few cases where it's safe to leave out the double quotes around a command substitution: in a variable assignment, there is an implicit pair of double quotes around the right-hand side, so new_filename=$(…) is equivalent to new_filename="$(…)". Note that this does not extend to export VARIABLE="$(value)", where the double quotes are necessary.

We now turn to your question about the use of sed. It is not necessary here; you can perform this substitution in bash. Bash has a pattern replacement parameter substitution feature, but it's limited to a constant replacement text. Bash also has a way to extract substrings from regexp matches with =~, through the BASH_REMATCH variable. After the match, ${BASH_REMATCH[0]} contains the portion of the string matched by the regexp, ${BASH_REMATCH[1]} contains the portion matched by the first parenthesized group and so on.

mv $filename $new_filename

Again, double quotes.

A working script

One possibility is to extract the text to replace from the regexp match, then perform a string replacement on it. Since I'm not using ./*, the file name may begin with a -, so I take care to use -- on the call to mv to ensure that the file name isn't seen as an option.

for filename in *; do
  if [[ $filename =~ _[0-9]\. ]]; then
    from=${BASH_REMATCH[0]}
    mv -- "$filename" "${filename//$from/${from/_/_0}}"
  fi
done

Another possibility is to match the whole file name as a regexp and splice a 0 into the bits. Note that this will behave differently in a corner case: if there are several occurrences of _[0-9]\., the code above replaces the first occurrence, while this replaces the last occurrence.

for filename in *; do
  if [[ $filename =~ ^(.*_)([0-9]\..*)$ ]]; then
    mv -- "$filename" "${BASH_REMATCH[1]}0${BASH_REMATCH[2]}"
  fi
done