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Just to make the obvious comments:

Let $q = (1 - \sqrt{5})/2$. Then $F_n=((-q^{-1})^{n} - q^n)/\sqrt{5}$. So $$\sum \frac{1}{F_n} = \sqrt{5} \sum \frac{q^n}{1-(-1)^n q^{2n}}.$$

This looks kind of like the logarithmic derivative of the modular form $\prod (1-q^n)$, but it's not exactly that. In any case, it is unlikely that someone has evaluated this sort-of-a-modular-form at this particular quadratic irrational. The natural thinkg to do is to evaluate modular forms when $\tau$ is a quadratic irrational, where $q=e^{2 \pi i \tau}$. For example, people might know what this sum equals at $q=e^{- \pi \sqrt{163}}$.

I'd be pleasantly surprised if there is a known answer. Of course, the safe money is always to bet on "transcendental" when you don't see a reason to expect anything else.

Looks like I was too pessimistic. It is known to be irrational, see http://mathworld.wolfram.com/ReciprocalFibonacciConstant.html (thanks Qiaochu!). They do indeed describe this quantity in terms of theta functions, which are a type of modular forms. To my surprise, they are able to prove things about these modular forms at $q$, although not to answer this question. Anyway, it looks like this reference has as much information as you can hope for.

Let $q = (1 - \sqrt{5})/2$. Then $F_n=((-q^{-1})^{n} - q^n)/\sqrt{5}$. So $$\sum \frac{1}{F_n} = \sqrt{5} \sum \frac{q^n}{1-(-1)^n q^{2n}}.$$
This looks kind of like the logarithmic derivative of the modular form $\prod (1-q^n)$, but it's not exactly that. In any case, it is unlikely that someone has evaluated this sort-of-a-modular-form at this particular quadratic irrational. The natural thinkg to do is to evaluate modular forms when $\tau$ is a quadratic irrational, where $q=e^{2 \pi i \tau}$. For example, people might know what this sum equals at $q=e^{- \pi \sqrt{163}}$.