Step 1 (Formula / Governing Law): (0.5 Mark)

The property of the adjoint of a square matrix $A$ of order $n$ is given by:

$$|\text{adj } A| = |A|^{n-1}$$

Additionally, for any square matrix $A$, the determinant of its transpose is equal to the determinant of the matrix itself:

$$|A^T| = |A|$$

Step 2 (Step-by-Step Working): (0.25 Mark)

Given that $A$ is a square matrix of order $n = 3$ and $|\text{adj } A| = 8$. Substituting these values into the formula:

$$|A|^{3-1} = 8$$ $$|A|^2 = 8$$

Taking the square root on both sides:

$$|A| = \pm\sqrt{8}$$ $$|A| = \pm 2\sqrt{2}$$

Since the question implies a standard evaluation where the determinant is typically considered in the context of the magnitude or the principal value provided in the options, we evaluate $|A^T| = |A|$.

Step 3 (Final Answer): (0.25 Mark)

Comparing the result with the given options, the value is:

$$\boxed{|A^T| = \pm 2\sqrt{2}}$$

Given the options provided, the correct choice is (D) $2\sqrt{2}$ (considering the positive root as per standard objective question conventions).