You can extend the Fibonacci numbers to a continuous function: $$F(n)=\frac{(1+\sqrt5)^x+(1-\sqrt5)^x}{2^x\sqrt5}$$ and you could apply the mean value theorem to that function, but I don't think doing so would solve your problem or bring you any closer to a solution.
All the mean value theorem says is that, at some point somewhere between the first Fibonacci number (1) and the seventh (13), if you drew a line tangent to your graph of $F(x)$, the tangent line would have the same slope as the line connecting the points $(1, 1)$ and $(7, 13)$. It wouldn't promise that this point, wherever it is (on which point the Mean Value Theorem is also unhelpfully silent), was anywhere near equally distant between $F(1)$ and $F(7)$, nor would there be any reason to think the value of $F$ at this point was one of the Fibonnaci numbers.
The easier solution would be to know in advance what number was half way between 1 and 13 (Take the difference, divide by two: $(13-1)/2=6$, add that to your starting value of one to get seven) and from there just start rattling off Fibonacci numbers until you find the ones nearest 7 (This will be 5 on the low side, 8 on the high. Two answers.)
You can still do this, even with the exponential growth of the fibonacci numbers which isn't really a problem at all. However, if you're trying to use this version of the middle fibonacci number as a measure of central tendency, that might or might not be a problem depending on what exactly you're hoping to do. Is that what you were thinking of when you mentioned worries about the fibonacci numbers not growing linearly?
EDIT This metaphor might, with luck, clarify why -- despite $F$'s exponential growth -- you can still use the approach described in paragraph three to find the Fibonacci number that is closest to being equally distant between $F(1)$ and $F(7)$. Suppose two people are running a race. The track is labeled in increments of one meter, with the runners starting at the mark for one meter and ending at the thirteen meter mark (that is, the track is twelve meters long). The displacement function of one runner, measuring how far the runner has gone from the starting line, grows linearly. The other, against common sense and probably also against medical advice, has a displacement function which increases exponentially. Where is the middle of the track for the first runner? Where is the middle for the second? In the same place: at the seven meter mark, six meters from the starting line and six meters from the finish line. When they get there will be different (and in fact the linear runner will get to the half way point in half the time it takes him to run the race, which isn't true of non-linear functions) but "where is the middle of the track?" doesn't depend on how fast you run it.
In this case, what's the middle of these two fibonacci numbers is like asking what the middle of the track is. The "when?" question, where different growth rates get you different answers is "at what value of $n$ is $F$ halfway between its initial value $F(1)$ and its final value $F(7)$?" What value is halfway between the two isn't so mysterious, that's just the middle of the track.
EDIT 2 I also made an embarrassing mistake in finding the midpoint of 1 and 13 in the previous answers, which has now been fixed. The midpoint is seven and the fibonacci numbers 5 and 8 tie for the closest number.
