Answer: 672

Solution:

Put the circle in the complex plane with center $0,$ $A = -2,$ $B = 2,$ and $C_i = 2\omega^i$ for $i = 1, \ldots, 6,$ where $\omega = e^{i\pi/7}.$ Then $AC_i = 2\,|\omega^i + 1|$ and $BC_i = 2\,|\omega^i - 1|,$ so $$AC_i \cdot BC_i = 4\,\bigl|\omega^{2i} - 1\bigr|.$$

As $i$ runs over $1, \ldots, 6,$ the numbers $\omega^{2i} = e^{2\pi i \cdot i/7}$ run over all six nontrivial $7$th roots of unity $\zeta^j.$ Since $\prod_{j=1}^{6} (x - \zeta^j)$ $= 1 + x + \cdots + x^6,$ plugging in $x = 1$ gives $\prod_{j=1}^{6} \bigl|1 - \zeta^j\bigr| = 7.$ Therefore $$\begin{aligned} n &= \prod_{i=1}^{6} 4\,\bigl|\omega^{2i} - 1\bigr| \\ &= 4^6 \cdot 7 = 28672. \end{aligned}$$

The remainder when $n$ is divided by $1000$ is $672.$