Let me assume $p > (1 + \varepsilon)(2 \ln n/ n)^{1/2}$, this in particular includes the case of constant $p \in (0,1)$.
If $p > (1 + \varepsilon)(2 \ln n/ n)^{1/2}$ then w.h.p. the binomial random graph $G(n,p)$ has diameter at most $2$. Also, for $p = \omega( \ln n / n)$ then w.h.p. every vertex has degree $(1 \pm o(1)) p n$. Thus the balls centered at any vertex have size $1$, $(1 \pm o(1)) p n$, and $n$; if the radius of the ball is $0, 1$ or at least $2$ respectively.
It seems that the only cases to consider are: how many singletons are needed to cover a radius-$1$ ball; and how many radius-$1$ balls are needed to cover the whole vertex set. In the first case any radius-$1$ ball trivially needs $(1 \pm o(1))pn$ singletons to cover it.
In the second case we are equivalently asking for the size of a minimum dominating set in $G(n,p)$, i.e. the minimum size of a set $S \subseteq V(G)$ where every vertex not in $S$ has a neighbour in $S$. This is known (see, e.g. [Wieland, Ben; Godbole, Anant P., On the domination number of a random graph, Electron. J. Comb. 8, No. 1, Research paper R37, 13 p. (2001). ZBL0989.05108.]) to be, w.h.p., concentrated around $(1 + o(1))\log_q(n)$, where $q = 1/(1-p)$.
Then the doubling constant will depend on which of $(1 + o(1))\log_q(n)$ or $(1 + o(1))pn$ is maximum. For constant $p \in (0,1)$ the former term is logarithmic and the second one is linear, so the doubling constant is $(1 + o(1))pn$.