To corroborate David's answer, I will compute $e_{n}(A)$, mainly because my guess in the above comment I made seems to be wrong, and I cannot erase it. (I think I did not see the possibility of choosing other $r$ cyclic generators.) I need this computation for my own work, so please correct me if I am careless again!

We have

$$e_{n}(A) = \frac{\#\text{Surj}(\mathbb{Z}_{p}^{n}, A)}{\#\text{Hom}(\mathbb{Z}_{p}^{n}, A)},$$

which is the proportion of $\mathbb{Z}_{p}$-linear maps $\mathbb{Z}_{p}^{n} \rightarrow A$ that are surjective. We may assume

$$A = \mathbb{Z}_{p}/(p^{\lambda_{1}}) \oplus \cdots \oplus \mathbb{Z}_{p}/(p^{\lambda_{l}}).$$

Now, note that any given map $\phi : \mathbb{Z}_{p}^{n} \rightarrow A$, its surjectivity can be checked by considering the surjectivity mod $p$, by Nakayama lemma. Moreover, the map

$$\text{Hom}_{\mathbb{Z}_{p}}(\mathbb{Z}_{p}^{n}, A) \twoheadrightarrow \text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)$$

given by the mod $p$ reduction is a group homomorphism, so each of its fiber must have the same size. This implies that

$$e_{n}(A) = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)} = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})},$$

which is equal to $0$ if $n < l$ and to the proportion of $n \times l$ $\mathbb{F}_{p}$-matrices with rank $l$ if $n \geq l$. By usual counting, we get

$$e_{n}(A) = \frac{(p^{n} - 1)(p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{l-1})}{p^{nl}} \\ = \left(1 - \frac{1}{p^{n-l+1}}\right)\left(1 - \frac{1}{p^{n-l+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right).$$

I hope I did not make mistake and I can cite my own answer in my work!

To corroborate David's answer, I will compute $e_{n}(A)$, mainly because my guess in the above comment I made seems to be wrong, and I cannot erase it. (I think I did not see the possibility of choosing other $r$ cyclic generators.)

We have

$$e_{n}(A) = \frac{\#\text{Surj}(\mathbb{Z}_{p}^{n}, A)}{\#\text{Hom}(\mathbb{Z}_{p}^{n}, A)},$$

which is the proportion of $\mathbb{Z}_{p}$-linear maps $\mathbb{Z}_{p}^{n} \rightarrow A$ that are surjective. We may assume

$$A = \mathbb{Z}_{p}/(p^{\lambda_{1}}) \oplus \cdots \oplus \mathbb{Z}_{p}/(p^{\lambda_{l}}).$$

Now, note that any given map $\phi : \mathbb{Z}_{p}^{n} \rightarrow A$, its surjectivity can be checked by considering the surjectivity mod $p$, by Nakayama lemma. Moreover, the map

$$\text{Hom}_{\mathbb{Z}_{p}}(\mathbb{Z}_{p}^{n}, A) \twoheadrightarrow \text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)$$

given by the mod $p$ reduction is a group homomorphism, so each of its fiber must have the same size. This implies that

$$e_{n}(A) = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)} = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})},$$

which is equal to $0$ if $n < l$ and to the proportion of $n \times l$ $\mathbb{F}_{p}$-matrices with rank $l$ if $n \geq l$. When $n \geq l$, we get

$$e_{n}(A) = \frac{(p^{n} - 1)(p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{l-1})}{p^{nl}} \\ = \left(1 - \frac{1}{p^{n-l+1}}\right)\left(1 - \frac{1}{p^{n-l+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right).$$

To corroborate David's answer, I will compute $e_{n}(A)$, mainly because the answer in the above comment seems to be wrong, and I cannot erase it. (I think I did not see the possibility of choosing other $r$ cyclic generators.) I need this computation for my own work, so please correct me if I am careless again!

We have

$$e_{n}(A) = \frac{\#\text{Surj}(\mathbb{Z}_{p}^{n}, A)}{\#\text{Hom}(\mathbb{Z}_{p}^{n}, A)},$$

which is the proportion of $\mathbb{Z}_{p}$-linear maps $\mathbb{Z}_{p}^{n} \rightarrow A$ that are surjective. Since $\mathbb{Z}_{p}$ is a PID, we may assume

$$A = \mathbb{Z}_{p}/(p^{\lambda_{1}}) \oplus \cdots \oplus \mathbb{Z}_{p}/(p^{\lambda_{l}}).$$

Now, note that any given map $\phi : \mathbb{Z}_{p}^{n} \rightarrow A$, its surjectivity can be checked by considering the surjectivity mod $p$, by Nakayama lemma. Moreover, the map $\text{Hom}_{\mathbb{Z}_{p}}(\mathbb{Z}_{p}^{n}, A) \twoheadrightarrow \text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)$ given by the mod $p$ reduction is a group homomorphism, so each of its fiber must have the same size. This implies that

$$e_{n}(A) = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)} = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})},$$

which is equal to $0$ if $n < l$ and to the proportion of $n \times l$ $\mathbb{F}_{p}$-matrices with rank $l$ if $n \geq l$. By usual counting you get

$$e_{n}(A) = \frac{(p^{n} - 1)(p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{l-1})}{p^{nl}} \\ = \left(1 - \frac{1}{p^{n-l+1}}\right)\left(1 - \frac{1}{p^{n-l+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right).$$

I hope I did not make mistake and I can cite my own answer in my work!

To corroborate David's answer, I will compute $e_{n}(A)$, mainly because my guess in the above comment I made seems to be wrong, and I cannot erase it. (I think I did not see the possibility of choosing other $r$ cyclic generators.) I need this computation for my own work, so please correct me if I am careless again!

We have

$$e_{n}(A) = \frac{\#\text{Surj}(\mathbb{Z}_{p}^{n}, A)}{\#\text{Hom}(\mathbb{Z}_{p}^{n}, A)},$$

which is the proportion of $\mathbb{Z}_{p}$-linear maps $\mathbb{Z}_{p}^{n} \rightarrow A$ that are surjective. We may assume

$$A = \mathbb{Z}_{p}/(p^{\lambda_{1}}) \oplus \cdots \oplus \mathbb{Z}_{p}/(p^{\lambda_{l}}).$$

Now, note that any given map $\phi : \mathbb{Z}_{p}^{n} \rightarrow A$, its surjectivity can be checked by considering the surjectivity mod $p$, by Nakayama lemma. Moreover, the map

$$\text{Hom}_{\mathbb{Z}_{p}}(\mathbb{Z}_{p}^{n}, A) \twoheadrightarrow \text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)$$

given by the mod $p$ reduction is a group homomorphism, so each of its fiber must have the same size. This implies that

$$e_{n}(A) = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, A/pA)} = \frac{\#\text{Surj}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})}{\#\text{Hom}_{\mathbb{F}_{p}}(\mathbb{F}_{p}^{n}, \mathbb{F}_{p}^{l})},$$

which is equal to $0$ if $n < l$ and to the proportion of $n \times l$ $\mathbb{F}_{p}$-matrices with rank $l$ if $n \geq l$. By usual counting, we get

$$e_{n}(A) = \frac{(p^{n} - 1)(p^{n} - p)(p^{n} - p^{2}) \cdots (p^{n} - p^{l-1})}{p^{nl}} \\ = \left(1 - \frac{1}{p^{n-l+1}}\right)\left(1 - \frac{1}{p^{n-l+2}}\right) \cdots \left(1 - \frac{1}{p^{n}}\right).$$

I hope I did not make mistake and I can cite my own answer in my work!