The terms in the expansion of $(1+ax)^n$ can be found using the binomial theorem, which states that
$$(1+ax)^n = \sum_{k=0}^n \binom{n}{k} (ax)^k$$
where $\binom{n}{k}$ is the binomial coefficient.
To find the coefficient of the $x^3$ term, we need to find the value of $k$ that satisfies $k \geq 3$ and $n-k=3$. Solving $n-k=3$ for $k$, we get $k=n-3$. Thus, we need to find the coefficient of $(ax)^{n-3}$ in the expansion. Using the formula for the binomial coefficient, we have
$$\binom{n}{n-3} = \frac{n!}{(n-3)!(3)!} = \frac{n(n-1)(n-2)}{6}$$
Multiplying this by $a^{n-3}$ and simplifying, we get the coefficient of $x^3$:
$$c = \frac{n(n-1)(n-2)}{6} a^{n-3}$$
Therefore, in the given expression, $c=\boxed{840}$, since the first three terms of the expansion have been provided.
