We will use induction with respect to $n$. The case for $n = 1$ is obvious. We suppose that every good subset of $Q_{n-1}$ has at most $2\left(\frac{3}{2}\right)^{n-1}$ elements.

Let $A_0 = \{(x_1, ..., x_n) \in A : x_n \neq 0\}$. We define the subsets $A_1, A_2$ similarly, that is
$$A_1 = \{(x_1, ..., x_n) \in A : x_n \neq 1\}, \quad A_2 = \{(x_1, ..., x_n) \in A : x_n \neq 2\}.$$
Since $A$ is a good set and $A_0$ is its subset, it follows that $A_0$ is also good. It means that for every three elements of $A_0$ there exists a coordinate which is different for every two of them. This coordinate cannot be the last one because $0$ cannot be there. Therefore the set $A_0'$ produced from the elements of $A_0$ by deleting the last coordinate is a good subset of $Q_{n-1}$.

Moreover, we observe that, if $|A_0| \geq 3$, then $|A_0'| = |A_0|$.
In fact, if $|A_0'| \neq |A_0|$, then there would exist an element $a \in A_0'$ such that $x, y \in A_0$, where $x, y$ arise from $a$ by adding $1$ and $2$, respectively, as last coordinate. However, if then $z$ is any other element of $A_0$, it cannot have as last coordinate $0$, and so $x, y, z$ will not form a good triad, absurd. Hence, from the induction hypothesis we have
$$|A_0| \leq \max\{2, |A_0'|\} \leq 2\left(\frac{3}{2}\right)^{n-1}.$$
Similarly, we get that: $|A_1|, |A_2| \leq 2\left(\frac{3}{2}\right)^{n-1}$. Since every element of $A$ appears exactly in two of the sets $A_0, A_1, A_2$, we conclude:
$$|A| = \frac{1}{2}(|A_0| + |A_1| + |A_2|) \leq 2\left(\frac{3}{2}\right)^n.$$