Write the given ratio in fractional form and simplify the right side.

$$\frac{^8C_n}{^6C_{n-1}} = \frac{56}{16}$$

Simplify the ratio on the right side by dividing both numbers by their greatest common divisor, which is 8:

$$\frac{^8C_n}{^6C_{n-1}} = \frac{7}{2}$$

$$^8C_n = \frac{8!}{n!(8-n)!}$$ $$^6C_{n-1} = \frac{6!}{(n-1)!(6-(n-1))!} = \frac{6!}{(n-1)!(7-n)!}$$

$$\frac{\frac{8!}{n!(8-n)!}}{\frac{6!}{(n-1)!(7-n)!}} = \frac{7}{2}$$

Invert and multiply:

$$\frac{8!}{n!(8-n)!} \times \frac{(n-1)!(7-n)!}{6!} = \frac{7}{2}$$

Expand the factorials to cancel common terms:
$8! = 8 \times 7 \times 6!$
$n! = n \times (n-1)!$
$(8-n)! = (8-n) \times (7-n)!$
Substitute these into the equation:

$$\frac{8 \times 7 \times 6!}{n \times (n-1)! \times (8-n) \times (7-n)!} \times \frac{(n-1)!(7-n)!}{6!} = \frac{7}{2}$$

Cancel out $6!$, $(n-1)!$, and $(7-n)!$:

$$\frac{8 \times 7}{n(8-n)} = \frac{7}{2}$$ $$\frac{56}{n(8-n)} = \frac{7}{2}$$

Cross-multiply:

$$56 \times 2 = 7 \times n(8-n)$$ $$112 = 7(8n - n^2)$$

Divide both sides by 7:

$$\frac{112}{7} = 8n - n^2$$ $$16 = 8n - n^2$$

Rearrange into a standard quadratic equation:

$$n^2 - 8n + 16 = 0$$

This is a perfect square trinomial:

$$(n-4)^2 = 0$$

Take the square root of both sides:

$$n-4 = 0$$ $$n = 4$$

For $^8C_n$, $n$ must be between 0 and 8 (inclusive). $n=4$ satisfies this.
For $^6C_{n-1}$, $n-1$ must be between 0 and 6 (inclusive). If $n=4$, then $n-1=3$, which satisfies $0 \le 3 \le 6$.
Both conditions are met.

The value of $n$ is $\boxed{4}$.