1. Use the determinant formula for adjugate

For an $m \times m$ matrix $M$,

$$\det(\operatorname{adj} M)=\big(\det M\big)^{m-1}.$$

Here, the matrix is $3\times 3$, so

$$\det(\operatorname{adj} M)=\big(\det M\big)^2.$$

Also, for a scalar multiple:

$$\det(kA)=k^3\det(A) \quad \text{for a }3\times 3\text{ matrix}.$$

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2. First compute $\det(A)$

Given

$$A=\begin{bmatrix}2&1&0\\1&2&-1\\0&-1&2\end{bmatrix}.$$

Expand along the first row:

$$\det(A)=2\begin{vmatrix}2&-1\\-1&2\end{vmatrix}-1\begin{vmatrix}1&-1\\0&2\end{vmatrix}+0.$$

Now,

$$\begin{vmatrix}2&-1\\-1&2\end{vmatrix}=4-1=3,$$

and

$$\begin{vmatrix}1&-1\\0&2\end{vmatrix}=2.$$

So,

$$\det(A)=2(3)-1(2)=6-2=4.$$

Hence,

$$\det(2A)=2^3\det(A)=8\cdot 4=32.$$

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3. Apply adjugate repeatedly

Let

$$M_0=2A.$$

Then

$$\det(M_0)=32.$$

Now define

$$M_1=\operatorname{adj}(M_0),\quad M_2=\operatorname{adj}(M_1),\quad M_3=\operatorname{adj}(M_2).$$

We need

$$\det(M_3)=\left|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))\right|.$$

Since each time for a $3\times 3$ matrix,

$$\det(\operatorname{adj} M)=(\det M)^2,$$

we get:

• First adjugate:

$$\det(M_1)=32^2.$$

• Second adjugate:

$$\det(M_2)=(32^2)^2=32^4.$$

• Third adjugate:

$$\det(M_3)=(32^4)^2=32^8.$$

Thus,

$$\left|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(2A)))\right|=32^8.$$

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4. Express in the form $(16)^n$

Since

$$32=2^5, \qquad 16=2^4,$$

we have

$$32^8=(2^5)^8=2^{40}=(2^4)^{10}=16^{10}.$$

Therefore,

$$n=10.$$

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5. Check with options

Option $\boxed{\text{C}}$ is correct.