$2^{3 n}-7 n-1 \text { is divisible by }$

Let $f(n) = 2^{3n} - 7n - 1 = 8^n - 7n - 1$.

Using the binomial expansion, $8^n = (1 + 7)^n = 1 + ^{n}C_{1}(7) + ^{n}C_{2}(7^2) + ^{n}C_{3}(7^3) + \dots + ^{n}C_{n}(7^n)$.

Substituting this into the expression for $f(n)$:

$f(n) = (1 + 7n + ^{n}C_{2}(49) + ^{n}C_{3}(7^3) + \dots + 7^n) - 7n - 1$.

$f(n) = ^{n}C_{2}(49) + ^{n}C_{3}(7^3) + \dots + 7^n$.

Since every term in the expansion contains at least $49$ as a factor, $f(n)$ is divisible by $49$.

Answer: 49