1. We start by noting that the least common multiple (LCM) of the numbers $6^6$, $8^8$, and $k$ must be $12^{12}$. We need to find the values of $k$ such that this condition holds.

2. First, we express each number in terms of its prime factorization:
$$6^6 = (2 \cdot 3)^6 = 2^6 \cdot 3^6$$
$$8^8 = (2^3)^8 = 2^{24}$$
$$12^{12} = (2^2 \cdot 3)^12 = 2^{24} \cdot 3^{12}$$

3. Let $k$ be expressed in terms of its prime factors as $k = 2^a \cdot 3^b$, where $a$ and $b$ are non-negative integers.

4. The LCM of $6^6$, $8^8$, and $k$ is given by:
$$\text{LCM}(6^6, 8^8, k) = \text{LCM}(2^6 \cdot 3^6, 2^{24}, 2^a \cdot 3^b)$$

5. The LCM of several numbers is found by taking the highest power of each prime that appears in any of the factorizations. Therefore:
$$\text{LCM}(2^6 \cdot 3^6, 2^{24}, 2^a \cdot 3^b) = 2^{\max(6, 24, a)} \cdot 3^{\max(6, 0, b)}$$

6. For the LCM to be equal to $12^{12} = 2^{24} \cdot 3^{12}$, we must have:
$$2^{\max(6, 24, a)} = 2^{24} \quad \text{and} \quad 3^{\max(6, 0, b)} = 3^{12}$$

7. From $2^{\max(6, 24, a)} = 2^{24}$, it follows that:
$$\max(6, 24, a) = 24 \implies a \leq 24$$

8. From $3^{\max(6, 0, b)} = 3^{12}$, it follows that:
$$\max(6, 0, b) = 12 \implies b = 12$$

9. Therefore, $a$ can take any integer value from $0$ to $24$, inclusive. This gives us $25$ possible values for $a$.

10. Since $b$ must be exactly $12$, there are no other constraints on $b$.

Thus, there are $25$ possible values for $k$.

The final answer is $\boxed{25}$.