I've just realized I was being a little bit slow. I had already found on the internet that $n^{-2}\sum\limits_{k=1}^n\phi(k)$ is roughly $3/\pi^2$ and stupidly didn't notice that I could "differentiate" this to get exactly what I want. That is, $\sum\limits_{k=1}^N \phi(k)$ is about $3N^2/\pi^2$, so the difference between the sum to $N+M$ and the sum to $N$ is around $6NM/\pi^2$, from which it follows that the average value near $N$ is around $6N/\pi^2$, which is entirely consistent with the well-known fact that the probability that two random integers are coprime is $6/\pi^2$.

I'm adding this paragraph after Greg's comment. To argue that the probability that two random integers are coprime, you observe that the probability that they do not have $p$ as a common factor is $(1-1/p^2)$. If you take the product of that over all $p$ then you've got the reciprocal of the Euler product formula for $\zeta(2)$, or $1^{-2}+2^{-2}+\ldots= \pi^2/6$. It's not that hard to turn these formal arguments into a rigorous proof, since everything converges nicely.