The only possible $n$ is $4$.
We call those edges which are the sides of a base the base edges, and call the remaining edges the lateral edges. Let $s$ be the sum of all numbers surrounding each face, and let $x$ be the sum of all numbers on the lateral edges.

Firstly, we consider the sum of all numbers on the base edges. Since they are the sides of the two bases, this is equal to $2s$. Also, they are the numbers not used on the lateral edges, and hence this is equal to $1 + 2 + \cdots + 3n - x$. This yields
$$x + 2s = \frac{3n(3n + 1)}{2}. \qquad (1)$$
Secondly, we consider the sides of the lateral faces. Each base edge is a side of $1$ lateral face, while each lateral edge is a side of $2$ lateral faces. The total sum of numbers on all these edges is $1 + 2 + \cdots + 3n + x$. These edges together surround $n$ faces, and so the total sum is $ns$. It follows that
$$ns - x = \frac{3n(3n + 1)}{2}. \qquad (2)$$
By (1) and (2), we obtain
$$s = \frac{(x + 2s) + (ns - x)}{n + 2} = \frac{3n(3n + 1)}{n + 2},$$
$$x = \frac{3n(3n + 1)}{2} - 2s = \frac{3n(3n + 1)(n - 2)}{2(n + 2)}.$$
As $s \in \mathbb{Z}$, we have $n + 2 \mid 3n(3n + 1)$. This implies $n + 2 \mid 30$.

Next, we have the bound $x \le 3n + (3n - 1) + \cdots + (2n + 1)$ since there are only $n$ lateral edges. Now,
$$\begin{align*} \frac{3n(3n + 1)(n - 2)}{2(n + 2)} &\le \frac{(5n + 1)n}{2} \\ \Leftrightarrow \quad 3(3n + 1)(n - 2) &\le (5n + 1)(n + 2) \\ \Leftrightarrow \quad 2(2n^2 - 13n - 4) &\le 0. \end{align*}$$
Thus, we have $n \le 6$. Together with $n + 2 \mid 30$ and the trivial bound $n \ge 3$, we easily obtain the only possibilities $n = 3, 4$.

For $n = 3$, we have $s = 18$ and $x = 9$.
* If the numbers on the lateral edges are $1$, $2$, $6$, then the numbers surrounding the three lateral surfaces are $1$, $2$, $a$, $b$; $1$, $6$, $c$, $d$ and $2$, $6$, $e$, $f$ respectively. Since $a + b = s - 1 - 2 = 15$, we must have $a = 7$ and $b = 8$. But then $c + d = 18 - 1 - 6 = 11$ has no solution with $c, d \in \{3, 4, 5, 9\}$.

* If the numbers on the lateral edges are $1$, $3$, $5$, then the numbers surrounding the three lateral surfaces are $1$, $3$, $a$, $b$; $1$, $5$, $c$, $d$ and $3$, $5$, $e$, $f$ respectively. Since $a + b = 14$, we must have $a = 6$ and $b = 8$. But then $c + d = 12$ has no solution with $c, d \in \{2, 4, 7, 9\}$.

* If the numbers on the lateral edges are $2$, $3$, $4$, then the numbers surrounding the three lateral surfaces are $2$, $3$, $a$, $b$; $2$, $4$, $c$, $d$ and $3$, $4$, $e$, $f$ respectively. Since $a + b = 13$, $c + d = 12$ and $e + f = 11$, none of $a$, $b$, $c$, $d$, $e$, $f$ is equal to $1$, which is impossible.

Therefore, it is not possible to have $n = 3$.

For $n = 4$, we have $s = 26$ and $x = 26$. An example is given below.