KDF and number of iterations for GPG I gather, from reading the GPG manual, that symmetric encryption based on a password uses one of a variety of Key Derivation Functions (KDF). Although not explicitly mentioned I assume that PBKDF2 is the default as the other options appear to be legacy algorithms from the PGP standard.

However I can't find any information on the number of iterations that the KDF does by default of whether there is a way to change that value manually. I want to know this because it would have an impact on the required complexity for the password (e.g. number of words generated using the diceware method) to achieve a given level of security.
Use gpg --s2k-mode 3 --s2k-count N, where N is the number of iterations you want to use.  The manual page says the default is 65536, and you can use any number between 1024 and 65011712.

If you like to tweak the defaults, I suggest making this number as large as you can bear it, without introducing noticeable slowdown (e.g., ideal would be to make the passphrase mangling process take about a second or so).

See also this answer to a different question, and the following questions/answers on IT Security.StackExchange:


What is purpose of --s2k-* GnuPG options?
gpg --perfect-code
an answer to "gpg encryption security"
an answer to "Good file encryption tools"
Do OpenPGP/GnuPG apply slow hash to password when encrypting/decrypting key?