As mentioned in the comments, Binet's formula,
$$F_n=\frac1{\sqrt{5}}(\phi^n-(-\phi)^{-n})$$
where $\phi=\frac12(1+\sqrt 5)$ is the golden ratio, is a closed-form expression for the Fibonacci numbers. See this related question for a few proofs.
As for counting how many Fibonacci numbers are there that are less than or equal to a given number $n$, one can derive an estimate from Binet's formula. The second term in the formula can be ignored for large enough $n$, so
$$F_n\approx\frac{\phi^n}{\sqrt{5}}$$
Solving for $n$ here gives
$$n=\frac{\log\,F_n+\frac12\log\,5}{\log\,\phi}$$
Taking the floor of that gives a reasonable estimate; that is, the expression
$$\left\lfloor\frac{\log\,n+\frac12\log\,5}{\log\,\phi}\right\rfloor$$
can be used to estimate the number of Fibonacci numbers $\le n$. This is inaccurate for small $n$, but does better for large $n$.
It turns out that by adding a fudge term of $\frac12$ to $n$, the false positives of the previous formula disappear. (Well, at least in the range I tested.) Thus,
$$\left\lfloor\frac{\log\left(n+\frac12\right)+\frac12\log\,5}{\log\,\phi}\right\rfloor$$
gives better results.
Suppose that your word contains $i$ B's. There are then $\binom{25}{12-i}$ ways to select the other letters to use. Then there are $12!/i!$ ways of ordering the $12-i$ distinct letters and the $i$ copies of B. Thus, there are a total of
$$\frac{25!}{(12-i)!(13+i)!} \cdot \frac{12!}{i!}$$
words of this form.
We wish to find
$$\sum_{i=0}^{12} \frac{25! \cdot 12!}{(12-i)!(13+i)!i!}.$$
Unfortunately, such a sum does not have a simple closed form. You may find a (terrible) 'closed form' in terms of hypergeometric functions, but the most efficient way I can see to evaluate this sum is to just do it.
Best Answer
A lot of people have mentioned Binet's formula. But I suspect this is not the most practical way to compute the nth Fibonacci number for large n, because it requires either having a very accurate value of $\sqrt{5}$ and carrying around lots of decimal places (if you want to do floating-point arithmetic) or expanding large powers of $1+\sqrt{5}$ using the binomial formula. The latter comes out to writing the Fibonacci number as a sum of binomial coefficients.
The following formulas hold, though:$$F_{2n-1}=F_n^2+F_{n-1}^2$$$$F_{2n}=(2F_{n-1}+F_n)\cdot F_n$$which you can find derivations of in the Wikipedia article on Fibonacci numbers. This lets you find $F_k$, for any $k$ even or odd, in terms of two Fibonacci numbers with approximately half the index. The result is faster than Binet's formula.