In order for $c_{n} \neq 0$, we must have $\operatorname{gcd}(n, 210)=1$, so we need only consider such $n$. The number $n^{c_{n}}-1$ is divisible by 210 iff it is divisible by each of 2, 3, 5, and 7, and we can consider the order of $n$ modulo each modulus separately; $c_{n}$ will simply be the LCM of these orders. We can ignore the modulus 2 because order is always 1. For the other moduli, the sets of orders are $a \in\{1,2\} \bmod 3$, $b \in\{1,2,4,4\} \bmod 5$, $c \in\{1,2,3,3,6,6\} \bmod 7$. By the Chinese Remainder Theorem, each triplet of choices from these three multisets occurs for exactly one $n$ in the range $\{1,2, \ldots, 210\}$, so the answer we seek is the sum of $\operatorname{lcm}(a, b, c)$ over $a, b, c$ in the Cartesian product of these multisets. For $a=1$ this table of LCMs is as follows: $\begin{array}{ccccccc} & 1 & 2 & 3 & 3 & 6 & 6 \\ \hline 1 & 1 & 2 & 3 & 3 & 6 & 6 \\ 2 & 2 & 2 & 6 & 6 & 6 & 6 \\ 4 & 4 & 4 & 12 & 12 & 12 & 12 \\ 4 & 4 & 4 & 12 & 12 & 12 & 12 \end{array}$ which has a sum of $21+56+28+56=161$. The table for $a=2$ is identical except for the top row, where $1,3,3$ are replaced by $2,6,6$, and thus has a total sum of 7 more, or 168. So our answer is $161+168=329$.