Let me give the answer for $n=2$. 

Fix a point $p \in C$ and call $x_1, \, x_2, \, \Delta$ the divisor classes

$\begin{equation*}
\begin{split}
x_1 & := \{(p, \, x) \, | \, x \in C\} \\
x_2 & := \{(x, \, p) \, | \, x \in C \} \\    
\Delta & := \{(x, \, x) \, | \, x \in C \}. 
\end{split}
\end{equation*}
$ 

Let $C^{(2)}$  be the second symmetric product of $C$, namely the quotient of $C^2$ by the action of $\mathbb{Z}/2 \mathbb{Z}$ exchanging the two factors, and let $x$, $\delta$ be the images of $x_1$ (or $x_2$) and $\Delta$ in $C^{(2)}$, respectively. 

It is a classical result that for a general curve $C$ of positive genus  the Neron-Severi group of $C^{(2)}$ is generated by $x$ and $\delta$, hence a basis for the Neron-Severi group of $C^2$ is given by $$x_1, \, x_2, \, \Delta.$$
The same result holds for the Neron-Severi group of the $n$-fold symmetric product $C^{(n)}$, hence one shows that the Neron Severi group of $C^n$ is generated, for general $C$, by the coordinate divisors and by the classes of diagonals.     

**References.** 
A. Kouvidakis: Divisors on symmetric product of curves, *Trans. Amer. Math. Soc.* **337** (1993), 117-128.