assign variable from bash mulitline timer output

How can I assign a variable to this multiline bash countdown timer output:

seconds=10
    start="$(($(date +%s) + $seconds))"
while [ "$start" -ge `date +%s` ]; do 
    time="$(( $start - `date +%s` ))"
printf '%s\r' "$(date -u -d "@$time" +%S)"
done 

and display the results with an:

echo -e $(timer_output)

syntax all in one line?

such as:

echo -e Script will exit in $(timer_output) seconds.

what I have now:

echo -e "\e[5m\e[1;31mScript will exit in:\e[0m\e[25m"
    seconds=10
    start="$(($(date +%s) + $seconds))"
while [ "$start" -ge `date +%s` ]; do 
    time="$(( $start - `date +%s` ))"
printf '%s\r' "$(date -u -d "@$time" +%S)"
done 

Thank you Terrance for a more elegant solution. Here is how it turned out with some color and it executes cleanly:

seconds=10
for ((i=${seconds};i>=0;i--)); do printf "\r\e[5m\e[1;31mScript will exit in $i seconds...\e[0m\e[25m"; sleep 1; done

I am not sure what you mean by assign the timer_output variable in a line that is constant. It would be better to use a for loop with the printf and the \r in your line so that it keeps looping and printing the same line over and over until the time as been met.

Example

seconds=10
for ((i=${seconds};i>=0;i--)); do printf "\rScript will exit in $i seconds."; sleep 1; done