1. Given the equation:
$$\log_a \log_a \log_a x = \log_{a^2} \log_{a^2} \log_{a^2} x$$
where $a = 256$.

2. Rewrite the right-hand side using the change of base formula:
$$\log_{a^2} y = \frac{1}{2} \log_a y$$
Therefore,
$$\log_{a^2} \log_{a^2} \log_{a^2} x = \frac{1}{2} \log_a \left( \frac{1}{2} \log_a \left( \frac{1}{2} \log_a x \right) \right)$$

3. Simplify the expression:
$$\log_a \log_a \log_a x = \frac{1}{2} \log_a \left( \frac{1}{2} \log_a \left( \frac{1}{2} \log_a x \right) \right)$$

4. Bring the $\frac{1}{2}$ inside as a square root for the inner expression:
$$\log_a \log_a \log_a x = \log_a \sqrt{\left( \frac{1}{2} \log_a \left( \frac{1}{2} \log_a x \right) \right)}$$

5. Eliminate the outer $\log_a$:
$$\log_a \log_a x = \sqrt{\left( \frac{1}{2} \log_a \left( \frac{1}{2} \log_a x \right) \right)}$$

6. Square both sides to get:
$$(\log_a \log_a x)^2 = \left( \frac{1}{2} \log_a \left( \frac{1}{2} \log_a x \right) \right)$$

7. Re-express the right-hand side as a sum of logs:
$$(\log_a \log_a x)^2 = \frac{1}{2} \left( \log_a \frac{1}{2} + \log_a \log_a x \right)$$

8. Let $u = \log_a \log_a x$:
$$u^2 = \frac{1}{2} \left( \log_a \frac{1}{2} + u \right)$$

9. Since $a = 256 = 2^8$, we have:
$$\log_a \frac{1}{2} = \log_{2^8} 2^{-1} = \frac{-1}{8}$$

10. Substitute this value into the equation:
$$u^2 = \frac{1}{2} \left( \frac{-1}{8} + u \right)$$

11. Multiply by 2 to clear the fraction:
$$2u^2 = \frac{-1}{8} + u$$

12. Rearrange the equation:
$$2u^2 - u + \frac{1}{8} = 0$$

13. Use the quadratic formula to solve for $u$:
$$u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
where $a = 2$, $b = -1$, and $c = \frac{1}{8}$.

14. Substitute the values:
$$u = \frac{1 \pm \sqrt{1 - 4 \cdot 2 \cdot \frac{1}{8}}}{2 \cdot 2} = \frac{1 \pm \sqrt{1 - 1}}{4} = \frac{1 \pm 0}{4} = \frac{1}{4}$$

15. Now, we can determine $x$:
$$u = \log_a \log_a x$$
$$\frac{1}{4} = \log_a \log_a x$$
$$a^{\frac{1}{4}} = \log_a x$$

16. Using the fact that $a = 2^8$:
$$2^2 = \log_{2^8} x$$
$$4 = \log_{2^8} x$$
$$2^5 = \log_2 x$$
$$x = 2^{32}$$

The final answer is $\boxed{2^{32}}$.