If you have some information about the first two moments of the summands, you can get something useful.  For example, a fairly standard form of the Chernoff bound would say something like if 
$$
  X = X_1 + \cdots + X_n
$$
with all $X_i$ discrete, supported in $[-1,1]$, centered (i.e., $\mathbb{E}[X_i]=0$), and independent, then 
$$\operatorname{Pr}[X > \lambda\sigma]\le e^{-\lambda^2/4}$$
where $\sigma^2 = \sum_{i\in [n]} \operatorname{Var}[X_i]$ is the variance of $X$.  Here is [a proof by Van Vu][vu].

If, instead, you are willing to assume that $\operatorname{E}[X_i]\le 0$ and replace $\sigma^2$ with the sum of $\operatorname{E}[X_i^2]$, you'll get a reasonably nice looking bound by just tracing through the argument.

Also, this will explain why you can't just infer a general answer.  The natural units for tail bounds are standard deviations from the mean. In this case, if you move the mean away from zero, then $\operatorname{E}[X_i^2]$ isn't quite the variance any more.

[vu]: http://cseweb.ucsd.edu/~klevchen/techniques/chernoff.pdf