Find all prime numbers $p$ such that $p^3$ divides the determinant
\[\begin{vmatrix} 2^2 & 1 & 1 & \dots & 1\\1 & 3^2 & 1 & \dots & 1\\ 1 & 1 & 4^2 & & 1\\ \vdots & \vdots & & \ddots & \\1 & 1 & 1 & & (p+7)^2 \end{vmatrix}.\]
1. We start by considering the determinant of the given matrix:
   \[
   \begin{vmatrix}
   2^2 & 1 & 1 & \dots & 11 & 3^2 & 1 & \dots & 1 \\
   1 & 1 & 4^2 & & 1 \\
   \vdots & \vdots & & \ddots & 1 \\
   1 & 1 & 1 & & (p+7)^2 
   \end{vmatrix}
   \]
   We need to find all prime numbers \( p \) such that \( p^3 \) divides this determinant.

2. By induction, we can show that the determinant of a matrix of the form:
   \[
   \begin{vmatrix}
   1 + a_1 & 1 & 1 & \dots & 1 \\
   1 & 1 + a_2 & 1 & \dots & 1 \\
   \vdots & \vdots & \ddots & \vdots \\
   1 & 1 & 1 & \dots & 1 + a_n
   \end{vmatrix}
   \]
   is given by:
   \[
   a_1 a_2 \dots a_n \left(1 + \frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}\right)
   \]

3. Applying this result to our matrix, we have:
   \[
   \begin{vmatrix}
   2^2 & 1 & 1 & \dots & 11 & 3^2 & 1 & \dots & 1 \\
   1 & 1 & 4^2 & & 1 \\
   \vdots & \vdots & & \ddots & 1 \\
   1 & 1 & 1 & & (p+7)^2 
   \end{vmatrix}
   = (2^2 - 1)(3^2 - 1) \dots ((p+7)^2 - 1) \left(1 + \frac{1}{2^2 - 1} + \frac{1}{3^2 - 1} + \dots + \frac{1}{(p+7)^2 - 1}\right)
   \]

4. Simplifying the terms, we get:
   \[
   (2^2 - 1) = 3, \quad (3^2 - 1) = 8, \quad \dots, \quad ((p+7)^2 - 1) = (p+7-1)(p+7+1) = (p+6)(p+8)
   \]
   Therefore, the product becomes:
   \[
   3 \cdot 8 \cdot \dots \cdot (p+6)(p+8)
   \]

5. For the second factor, we use the identity:
   \[
   \frac{1}{k^2 - 1} = \frac{1}{2(k-1)} - \frac{1}{2(k+1)}
   \]
   This telescopes to:
   \[
   \frac{1}{2} \left(2 + 1 - \frac{1}{3} + \frac{1}{2} - \frac{1}{4} + \dots + \frac{1}{p+6} - \frac{1}{p+8}\right) = \frac{1}{2} \left(\frac{7}{2} - \frac{1}{p+7} - \frac{1}{p+8}\right)
   \]

6. Combining these results, the determinant is:
   \[
   3 \cdot 8 \cdot \dots \cdot (p+6)(p+8) \left(\frac{7}{2} - \frac{1}{p+7} - \frac{1}{p+8}\right)
   \]

7. We need to check when \( p^3 \) divides this expression. For \( p = 2, 3, 5 \), we can check by hand:
   - For \( p = 2 \):
     \[
     3 \cdot 8 \cdot \dots \cdot 8 \left(\frac{7}{2} - \frac{1}{9} - \frac{1}{10}\right)
     \]
     This is divisible by \( 2^3 \).

   - For \( p = 3 \):
     \[
     3 \cdot 8 \cdot \dots \cdot 10 \left(\frac{7}{2} - \frac{1}{10} - \frac{1}{11}\right)
     \]
     This is divisible by \( 3^3 \).

   - For \( p = 5 \):
     \[
     3 \cdot 8 \cdot \dots \cdot 12 \left(\frac{7}{2} - \frac{1}{12} - \frac{1}{13}\right)
     \]
     This is divisible by \( 5^3 \).

8. For \( p > 7 \), the first product is divisible exactly by \( p^2 \). We need to check when the last factor is divisible by \( p \). The fractions are invertible modulo \( p \), so the factor equals:
   \[
   \frac{7}{2} - \frac{1}{7} - \frac{1}{8} = \frac{181}{56} \mod p
   \]
   Therefore, the final answer is:
   \[
   \boxed{p \in \{2, 3, 5, 181\}}
   \]