Bases

We are given that $n = \overline{ABC}_8$ and $n = \overline{CBA}_6$, where the subscripts represent the bases. Let's convert both representations to base $10$ and set them equal to find the value of $n$.

In base $8$, the number $n$ can be represented as:
\[n = A \cdot 8^0 + B \cdot 8^1 + C \cdot 8^2.\]

In base $6$, the number $n$ can be represented as:
\[n = C \cdot 6^0 + B \cdot 6^1 + A \cdot 6^2.\]

Equating the two expressions for $n$:
\[A + 8B + 64C = C + 6B + 36A.\]

Simplify the equation:
\[35A + 2B = 56C.\]

Since $A$, $B$, and $C$ are all non-negative integers, we can observe that the maximum value for $C$ is $1$ (since the left side should not exceed the right side by too much). Substituting $C = 1$ gives:
\[35A + 2B = 56.\]

Now, we need to find non-negative integer values of $A$ and $B$ that satisfy this equation. The largest possible value for $A$ is $1$, and for $B$ is $17$. When $A = 1$ and $B = 17$, the equation is satisfied.

So, the largest possible value of $n$ in base $10$ is obtained by substituting $A = 1$, $B = 17$, and $C = 1$ into the base $8$ representation:
\[n = 1 \cdot 8^0 + 17 \cdot 8^1 + 1 \cdot 8^2 = 1 + 136 + 64 = \boxed{201}.\]