For any group $G$ and any finite $G$-set $X$, write $a(X)$ for the average cardinality of a $G$-orbit in $X$, that is, with obvious notation $$ a(X)=\frac{\text{number of points}}{\text{number of orbits}}=\frac{|X|}{|G\backslash X|}\ . $$ Let $G$ be a (necessarily finite) group, $X$ a finite faithful $G$-set, $P(X)$ the power set of $X$, and let us define the iterated power sets $P^2(X)$, $P^3(X)$ of $X$ in the obvious way.
Consider the following condition on our $G$-set $X$:
Condition. $a(P^k(X))$ tends to $|G|$ (the order of $G$) as $k$ tends to $+\infty$.
Do all finite faithful $G$-sets satisfy this condition?
Yes, finite faithful $G$-sets are asymptotically free.
To see this, let $g$ be the order of $G$, say that a point of a $G$-set is $G$-free (or just free if $G$ is obvious form the context) if its stabilizer in $G$ is trivial, and consider the condition:
$(1)$ The proportion of free point in $P^k(X)$ tends to $1$ as $k$ tends to $\infty$.
We claim that $(1)$ is equivalent to the condition in the question, which is
$(2)$ The average number of points by orbit in $P^k(X)$ tends to $g$ as $k$ tends to $\infty$.
To translate $(1)$ and $(2)$ into precise mathematical statements, let $n(k)$ be the number of points of $P^k(X)$, let $f(k)$ be the number of free orbits (so that the number of free points is $gf(k)$), and let $r(k)$ be the number of orbits. Then $(1)$ says that $gf(k)/n(k)$ tends to $1$ as $k$ tends to $\infty$, and $(2)$ says that $n(k)/r(k)$ tends to $g$ as $k$ tends to $\infty$.
We'll prove
$(3)$ Conditions $(1)$ and $(2)$ are equivalent.
$(4)$ Condition $(2)$ holds when $g$ is a prime number $p$.
Let's prove that all finite faithful $G$-sets satisfy $(1)$ and $(2)$ taking $(3)$ and $(4)$ for granted. (The references to $(3)$ will be implicit.)
Proof. We can assume $G\ne\{1\}$. Let $\mathcal H'$ be the set of all subgroups $H$ of $G$ such that $H\ne\{1\}$, and let $\mathcal H$ be the set of all minimal elements of $\mathcal H'$. In particular $\mathcal H$ is nonempty and any $H\in\mathcal H$ has prime order.
Let $\xi\in P^k(X)$. Then $\xi$ is $G$-free if and only if it is $H$-free for all $H\in\mathcal H$.
For each subgroup $K$ of $G$ let $F(K,k)$ be the set of $K$-free points in $P^k(X)$. Then $F(G,k)$ is the intersection of the finitely many sets $F(H,k)$ with $H\in\mathcal H$.
As the proportion of points of $P^k(X)$ which are in $F(H,k)$ with $H\in\mathcal H$ tends to $1$ by $(4)$, this property also holds for $F(G,k)$, that is, $(1)$ holds. $\square$
We're left with proving $(3)$ and $(4)$.
Proof of $(3)$. (Recall that $(3)$ says that $(1)$ and $(2)$ are equivalent.) We have, in the notation introduced just before the statement of $(3)$, $$ f(k)+\frac{n(k)-gf(k)}{g/2}\le r(k)\le f(k)+ n(k)-gf(k), $$ that is $$ \frac2g-\frac{f(k)}{n(k)}\le\frac{r(k)}{n(k)}\le1-(g-1)\ \frac{f(k)}{n(k)}\ . $$ Thus, if $f(k)/n(k)$ tends to $1/g$, so does $r(k)/n(k)$. Assume conversely that $r(k)/n(k)$ tends to $1/g$, and let $\varepsilon$ be positive. For $k$ large enough we have $$ \frac2g-\frac{f(k)}{n(k)}<\frac1g+\varepsilon, $$ that is $$ 0\le\frac1g-\frac{f(k)}{n(k)}<\varepsilon. $$ This shows that $f(k)/n(k)$ tends to $1/g$. $\square$
Proof of $(4)$. (Recall that $(4)$ says that $(2)$ holds when $g$ is a prime number $p$.) Writing $m(k)$ for the number of fixed points of $G$ in $P^k(X)$, we get $$ n(k)=pf(k)+m(k),\quad r(k)=f(k)+m(k), $$ and we want to show that $r(k)/n(k)$ tends to $1/p$. As we have $$ \frac{pr(k)}{n(k)}=1+(p-1)\ \frac{m(k)}{n(k)}\ , $$ it suffices to verify that $m(k)/n(k)$ tends to $0$. Note that $$ \frac{m(k+1)}{n(k+1)}=\frac{2^{r(k)}}{2^{n(k)}}=2^{r(k)-n(k)}, $$ and we only need to verify that $$ \frac{p}{p-1}\ \Big(n(k+1)-r(k+1)\Big)=n(k+1)-m(k+1) $$ $$ =2^{n(k)}-2^{r(k)}=2^{r(k)}\left(2^{n(k)-r(k)}-1\right) $$ tends to $\infty$. But we have $n(k)-r(k)\ge1$ by faithfulness and the inequality $pr(k)\ge n(k)$ implies that $r(k)$ tends to $\infty$. $\square$