1. Convert the problem to the complex plane:
- Let $\omega = \operatorname{cis}\left(\frac{2\pi}{n}\right)$, where $\operatorname{cis}(\theta) = \cos(\theta) + i\sin(\theta)$.
- The vertices of the regular $n$-gon can be represented as $A_j^{(0)} = \omega^j$ for $j = 0, 1, \ldots, n-1$.

2. Determine the centroid of the triangle:
- The centroid of a triangle with vertices $\alpha, \beta, \gamma$ is given by $\frac{\alpha + \beta + \gamma}{3}$.
- For $i = 1$, the centroid $A_j^{(1)}$ is:
$$A_j^{(1)} = \frac{\omega^j + \omega^{j+3^6} + \omega^{j+2 \cdot 3^6}}{3}$$

3. Generalize the centroid calculation:
- For $i = 2$, the centroid $A_j^{(2)}$ is:
$$A_j^{(2)} = \frac{\omega^j + \omega^{j+3^5} + \omega^{j+2 \cdot 3^5} + \ldots + \omega^{j+8 \cdot 3^5}}{9}$$
- By induction, for $i = 7$, the centroid $A_j^{(7)}$ is:
$$A_j^{(7)} = \frac{\omega^j + \omega^{j+1} + \omega^{j+2} + \ldots + \omega^{j+3^7-1}}{3^7}$$

4. Simplify the expression:
- Notice that the sum of all $n$-th roots of unity is zero:
$$\omega^0 + \omega^1 + \ldots + \omega^{n-1} = 0$$
- Therefore, the sum $\omega^j + \omega^{j+1} + \ldots + \omega^{j+3^7-1}$ is a subset of the $n$-th roots of unity, and it sums to zero:
$$\omega^j + \omega^{j+1} + \ldots + \omega^{j+3^7-1} = 0$$

5. Calculate the magnitude:
- Since the sum is zero, we have:
$$A_j^{(7)} = \frac{0 - \omega^{j-1}}{3^7} = -\frac{\omega^{j-1}}{3^7}$$
- The magnitude of $A_j^{(7)}$ is:
$$|A_j^{(7)}| = \left| -\frac{\omega^{j-1}}{3^7} \right| = \frac{1}{3^7}$$

6. Compare magnitudes:
- Initially, $|OA_{2014}^{(0)}| = 1$.
- Therefore, the ratio is:
$$\frac{|OA_{2014}^{(7)}|}{|OA_{2014}^{(0)}|} = \frac{\frac{1}{3^7}}{1} = \frac{1}{3^7}$$
- Since $3^7 = 2187$, we have:
$$\frac{1}{3^7} = \frac{1}{2187}$$

7. **Find $p$ and $q$:**
- Here, $p = 1$ and $q = 2187$, so $p + q = 1 + 2187 = 2188$.

The final answer is $\boxed{2188}$