No one has provided an exact answer, so I'll discuss approximations. If you're content with an approximate answer, you could consider
I. Use simulation to estimate the fraction of all possible length j passwords composed of the 68 allowed characters (26 letters, 10 numbers, 32 special characters), i.e. $68^j$, which are in compliance with the rules. Do separately for j = 14, 13, etc. Multiply the fraction in compliance by the number of passwords of that length, and sum over lengths of 6 to 14. There might also be some clever ways to incorporate some of the rules restrictions into the simulated candidates, then estimate the fraction of those which are in compliance with the remaining rules.
II. An upper bound approximation (code strands are in MATLAB, but they should be understandable by all):
There are 68 allowed characters, so if no passwords were disallowed, there would be
sum(68.^(6:14)) = 4.587e25 possible passwords.
For a length j password, there is
$P$(no numbers) + $P$(no letters) - $P$(no numbers or letters) =
$(58/68)^j+(42/68)^j-(32/68)^j$ probability of not having at least one number and one letter. The probability of not having at least one number is dominant within this.
That comes out as follows:
j P(not having at least one letter and one number in password of length j)
So, decrementing the total number of passwords of lengths 6 to 14 only by those disallowed due to not having at least one number and one letter, results in a total of
= 4.086e+25 passwords.
The probability of not having 9 or more numbers given at least one exceeds 0.99996. I believe that disallowance due to all other reasons would only contribute a "very small" decrement, although I have not quantified that. Therefore, the number of allowed passwords will be "a little less" than 4.086e+25, if I haven't made a mistake. I leave any further refinement to others. But this is probably "good enough for government use".