Let D equal the number of digits for the security code. Let K equal the number of digits on the keypad.
For a binary keypad (0, 1) and a three-digit code, our set consists of (000E, 001E, 010E, 011E, 100E, 101E, 110E, 111E). Thus, we have at most $2^3 \times (3+1)$ possible keys to enter, where the plus one is for the ENTER key. Generally speaking, we'd have $K^D \times (D+1)$ possible keys to enter.
Without the ENTER key, we could keep pressing the keypad until all combinations have formed. The minimum number of keys required would be $K^D + D$.
So for a binary keypad, we'd have:
|D||String||Keys (wo/ E)||Keys (w/ E)|
Thus, one only needs to know the string concatenations to be able to guess the right combination if no ENTER or code-stopper key is required.
Update: See De Bruijn Sequence and ProjectEuler #265 for a related problem.