How to get the second last argument from shell script?

In bash/ksh/zsh you can simply ${@: -2:1}

$ set a b c d 
$ echo ${@: -1:1}
c

In POSIX sh you can use eval:

$ set a b c d 
$ echo $(eval "echo \$$(($#-2))")
c

set -- "first argument" "second argument" \
       "third argument" "fourth argument" \
       "fifth argument"
second_to_last="${@:(-2):1}"
echo "$second_to_last"

Note the quoting, which ensures that arguments with whitespace stick together -- which your original solution doesn't do.

n=$(($#-1))
second_to_last=${!n}
echo "$second_to_last"

There are some options for all bash versions:

$ set -- aa bb cc dd ee ff
$ echo "${@: -2:1}   ${@:(-2):1}   ${@:(~1):1}   ${@:~1:1}   ${@:$#-1:1}"
ee   ee   ee   ee   ee

The (~) is the bitwise negation operator (search in the ARITHMETIC EVALUATION section).
It means flip all bits.

The selection even could be done with (integer) variables:

$ a=1  ; b=-a; echo "${@:b-1:1}   ${@:(b-1):1}   ${@:(~a):1}   ${@:~a:1}   ${@:$#-a:1}"
ee   ee   ee   ee   ee

$ a=2  ; b=-a; echo "${@:b-1:1}   ${@:(b-1):1}   ${@:(~a):1}   ${@:~a:1}   ${@:$#-a:1}"
dd   dd   dd   dd   dd

For really old shells, you must use eval:

eval "printf \"%s\n\" \"\$$(($#-1))\""