12. The rules of a "level-passing game" stipulate: on the $n$th level, a die must be rolled $n$ times, and if the sum of the points that appear in these $n$ rolls is greater than $2^{n}$, it counts as passing the level. Therefore, the probability of consecutively passing the first 3 levels is $\qquad$
12. $\frac{100}{243}$.

Since the dice are uniform cubes, the probability of each number appearing after a throw is equal.

Let event $A_{n}$ be “failure at the $n$-th level”, then the complementary event $B_{n}$ is “success at the $n$-th level”. In the $n$-th game, the total number of basic events is $6^{n}$.

First Level: The number of basic events in event $A_{1}$ is 2 (i.e., the numbers 1 and 2 appear), so the probability of passing this level is
$$
P_{B_{1}}=1-P_{A_{1}}=1-\frac{2}{6}=\frac{2}{3} \text {. }
$$

Second Level: The number of basic events in event $A_{2}$ is the sum of the positive integer solutions of the equation $x+y=a$ when $a$ takes the values $2, 3, 4$, which is 6. So, the probability of passing this level is
$$
P_{B_{2}}=1-P_{A_{2}}=1-\frac{6}{6^{2}}=\frac{5}{6} \text {. }
$$

Third Level: The number of basic events in event $A_{3}$ is the sum of the positive integer solutions of the equation $x+y+z=a$ when $a$ takes the values $3, 4, 5, 6, 7, 8$, which is 56. So, the probability of passing this level is
$$
P_{B_{3}}=1-P_{A_{3}}=1-\frac{56}{6^{3}}=\frac{20}{27} \text {. }
$$

Therefore, the probability of passing all three levels is
$$
P_{B_{1}} \times P_{B_{2}} \times P_{B_{3}}=\frac{100}{243} .
$$