I think the answer is yes. Let $X=\mathbb Z_{p_1}\times \ldots \times \mathbb Z_{p_k}$. Notice that the metric on $X$ is an ultrametric, so that if $A$ is a subset of $X$ of diameter $d$ and $a\in A$, then the ball of radius $d$ about $a$ is a superset of $A$ that has the same diameter. Denote this ball by $B(A)$. 

Writing $a=(a_1,\ldots,a_k)$ in $X$. The ball of radius $r$ about $a$, $B_r(a)$, is the product $B_r(a_1)\times\ldots\times B_r(a_k)$, where $B_r(a_i)$ is the ball of radius $r$ about $a_i$ in $\mathbb Z_{p_i}$. Of course $B_r(a_i)$ is equal to $B_{p_i^{-s_i}}(a_i)$ where $s_i$ is the largest integer such that $p_i^{-s_i}\le r$. The diameter of the ball is $d=\max_i p_i^{-s_i}$. The volume of the ball with respect to Haar measure is $\prod_i p_i^{-s_i}\le d^k$. In general, we have for any ball, $\mu(B)\le \text{diam}(B)^k$. 

Now if $A_1,\ldots,A_n$ is a cover of $X$, then so is $B(A_1),\ldots,B(A_n)$. These sets have the same diameters by the first observation. Now we have
$$
\sum\text{diam}(A_i)^k = \sum\text{diam}(B(A_i))^k \ge \sum \mu(B(A_i)) \ge 1.
$$

Hence $c\ge 1$.

On the other hand, by considering the map $\phi\colon t\mapsto (t\bmod \log p_1,\ldots,t\bmod\log p_k)$ (using density of $\phi(\mathbb R)$ in the $k$-dimensional torus), we can find a sequence $t_n\to\infty$ such that $\phi(t_n)$ is a vector with all coordinates lying in $(\log p_j-\epsilon_n,\log p_j)$ with $\epsilon_n\to 0$. Write $m^i_n=\lceil t_n\log p_i\rceil$. Then $e^{-t_n}$ lies between $p_i^{-m^i_n}$ and $p_i^{-m^i_n}e^{\epsilon_n}$. 

For balls of diameter $e^{-t_n}$, the measure of the ball is close (within a factor $e^{k\epsilon_n}$) to the $k$th power of the diameter. Hence $c=1$.