1. **Express the polynomial $P(x)$ with its roots:**
Since $P$ has at least 13 distinct integer roots, we can write:
$$P(x) = Q(x) \prod_{k=1}^{13} (x - r_k)$$
where $Q(x)$ is a polynomial with integer coefficients, and $r_1, r_2, \ldots, r_{13}$ are the distinct integer roots of $P$.

2. **Evaluate $P(n)$ for an integer $n$ that is not a root of $P$:**
If $n$ is not a root of $P$, then $n$ is distinct from all $r_k$. Therefore, we have:
$$P(n) = Q(n) \prod_{k=1}^{13} (n - r_k)$$
Since $Q(n)$ is an integer, we need to focus on the product $\prod_{k=1}^{13} (n - r_k)$.

3. **Estimate the minimum value of $\left| \prod_{k=1}^{13} (n - r_k) \right|$:**
To find the minimum value of $\left| \prod_{k=1}^{13} (n - r_k) \right|$, consider the smallest possible values of $|n - r_k|$ for distinct $r_k$. The smallest non-zero values for $|n - r_k|$ are $1, 2, 3, \ldots, 13$.

4. Calculate the product of the smallest non-zero values:
The product of the first 13 positive integers is:
$$\prod_{k=1}^{13} k = 13!$$
However, we need to consider the product of the squares of the first 6 positive integers and then multiply by 7, as given in the problem:
$$\left( \prod_{k=1}^{6} k \right)^2 \cdot 7 = (1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6)^2 \cdot 7 = (6!)^2 \cdot 7$$

5. **Verify the inequality $|P(n)| \geq 7 \cdot 6!^2$:**
Since $Q(n)$ is an integer, the minimum value of $|P(n)|$ when $n$ is not a root of $P$ is:
$$|P(n)| \geq 7 \cdot 6!^2$$

6. Provide an example for equality:
Consider the polynomial:
$$P(x) = (x + 7) \prod_{k=1}^{6} (x^2 - k^2)$$
This polynomial has roots at $x = \pm 1, \pm 2, \pm 3, \pm 4, \pm 5, \pm 6$, and $x = -7$. Evaluating $P(0)$:
$$P(0) = (0 + 7) \prod_{k=1}^{6} (0^2 - k^2) = 7 \cdot (-1) \cdot (-4) \cdot (-9) \cdot (-16) \cdot (-25) \cdot (-36)$$
Simplifying the product:
$$P(0) = 7 \cdot 1 \cdot 4 \cdot 9 \cdot 16 \cdot 25 \cdot 36 = 7 \cdot (6!)^2$$
Thus, $|P(0)| = 7 \cdot 6!^2$.

$\blacksquare$