It's worth distinguishing between the prime coefficients $a_p$, and the coefficients $a_n$ for general $n$. Let's look at $a_p$ first.

Firstly: for elliptic curves, it is fairly easy and elementary to prove that $|a_p| < 2\sqrt{p}$ (Hasse's inequality). This should be in any decent textbook e.g. Silverman. And this is the best bound you're going to get, because for any fixed elliptic curve the set $\{ a_p / (2\sqrt{p}) : p\ prime\}$ is dense in $(-1, 1)$, and in fact obeys a very specific distribution. (This distribution depends on whether or not E has complex multiplication. If E does not have CM, it the *Sato-Tate distribution*, a long-standing conjecture recently settled by Barnet-Lamb, Geraghty, Harris and Taylor. In the CM case you get a different, simpler distribution; thanks KConrad for pointing this out.)

For general modular eigenforms of weight $k$, the "right" bound is $|a_p| \le 2p^{(k-1)/2}$, but this is a very deep theorem (it follows from Deligne's work on the Weil conjectures). There is an easy elementary argument that gives $a_p = O(p^{k/2})$: this is in Miyake's book (corollary 2.1.6 if I recall correctly). By purely analytic methods you can push this a bit further, e.g. Rankin proved $a_p = O(p^{k/2 - 1/5})$ if I remember correctly, but you can't get the "right" bound this way.

For general $n$, the relations giving the $a_n$ in terms of the $a_p$ mean that the Deligne bound $|a_p| \le 2p^{(k-1)/2}$ turns into something like $a_n \le n^{(k-1)/2} d(n)$, where $d(n)$ is the number of divisors of $n$; this is $O(n^{(k-1)/2 + \epsilon})$ for any $\epsilon > 0$ (but it is *not* $O(n^{(k-1)/2})$).

(EDIT: For Hasse's inequality it's obvious that we have strict inequality $|a_p| < 2\sqrt{p}$ simply because $a_p \in \mathbf{Z}$ and $2\sqrt{p}$ isn't. But for general modular forms one only gets the non-strict inequality $|a_p| \le 2p^{(k-1)/2}$, not a strict inequality as I originally claimed (thanks to François for pointing this out); strict inequality is conjectured to hold if $k > 1$, and Coleman and Edixhoven have shown that this would follow from some standard conjectures on varieties over finite fields. If $k = 1$ then in fact $a_p = 2$ for a positive density set of primes, so equality definitely can occur in this case.)

(EDIT: Just to emphasize, if you have an elliptic curve over $\mathbf{Q}$, the Hasse bound $|a_p| < 2\sqrt{p}$ "follows from" the Deligne bound and the fact that $E$ is modular, but this would be a ridiculously laborious way of proving that: the direct elementary proof of Hasse's inequality is vastly easier than using modularity. In fact (some parts of) Deligne's proof can be interpreted as "trying to adapt Hasse's proof to a general algebraic variety", so the flow of information here is the other way.)