Adding a column with random values to end of CSV

You have the syntax of your while...do...done loop wrong. You are running this:

while read x ; OneTimePass=$(openssl rand -base64 14 | head -c 6) ; do ...

The while keyword's expected format is explained in help while:

$ help while
while: while COMMANDS; do COMMANDS; done
    Execute commands as long as a test succeeds.
    
    Expand and execute COMMANDS as long as the final command in the
    `while' COMMANDS has an exit status of zero.
    
    Exit Status:
    Returns the status of the last command executed.

Here, you are giving it two commands: read x ; and OneTimePass=$(openssl rand -base64 14 | head -c 6), and the second command always works, it doesn't have any ending since you can always re-run the openssl command. This is why the while loop never exits and you just keep writing more and more lines. What you're after is this:

while read x; do
    something
done

Here's a working version of your script with some other improvements such as proper quoting and avoiding CAPS for your variable names:

#!/bin/bash

#add column to csv
orig_file="new-users2.csv"
new_file="Output.csv"

printf "%s,%s\n" "$(head -1 "$orig_file")" "One Time Password" > "$new_file"
tail -n +2 "$orig_file" |
  while read -r x; do
    OneTimePass="$(openssl rand -base64 14 | head -c 6)"
    printf '%s,%s\n' "$x" "$OneTimePass"
  done >> "$new_file"

Personally, I would avoid hardcoding the file names since that makes the script harder to use and less versatile. I would instead take the input file name as an argument, and then print to standard output so you can choose whatever output file you want:

#!/bin/bash

#add column to csv
orig_file=$1

printf "%s,%s\n" "$(head -1 "$orig_file")" "One Time Password"
tail -n +2 "$orig_file" |
  while read -r x; do
    OneTimePass="$(openssl rand -base64 14 | head -c 6)"
    printf '%s,%s\n' "$x" "$OneTimePass"
  done

You can then run it like this:

foo.sh new-users2.csv > Output.csv 

I tested with this input file:

$ cat new-users2.csv 
name,age
Bob,45
Alice,36

Which results in:

$ foo.sh new-users2.csv 
name,age,One Time Password
Bob,45,BTkLQW
Alice,36,CzQa4U

Using Miller, and cheekily borrowing @terdon's example file

$ mlr --csv put -S '
    $["One Time Password"] = substr(system("openssl rand -base64 5"),0,5)
' new-users2.csv 
name,age,One Time Password
Bob,45,kIrlrl
Alice,36,h1OBp3

(I changed the openssl rand parameter from 14 to 5 bytes since you should need no more than 36 bits to generate six base64 characters. Change it back if it is important for some reason.)

You can do a similar thing with awk's system() function but handling of the CSV structure and headers is messier. Perhaps neater with awk would be to use getline ex.:

$ awk -F, '
    BEGIN {OFS=FS; cmd = "openssl rand -base64 5"} 
    NR==1 {$(NF+1) = "One Time Password"} 
    NR >1 {cmd | getline var; close(cmd); $(NF+1) = substr(var,1,6)} 
    1
' new-users2.csv
name,age,One Time Password
Bob,45,CovLkz
Alice,36,PgLbD4

Unlike the Miller version, this will only work for "simple" CSV (it won't handle embedded "," characters within fields for example).