The Burnside $\mathbb Q$-algebra $\mathbb QB(G)$ of a group $G$ is usually considered only when $G$ is finite; see Section 3.1 of the text
[1] Serge Bouc, https://pdfs.semanticscholar.org/aff3/8005a0c57f36a15b01747616e738f1f9eb4d.pdf
Moreover, we have in this case $\mathbb QB(G)\simeq\mathbb Q^n$ with $n=\dim_{\mathbb Q}\mathbb QB(G)$, and thus $\mathbb QB(G)\simeq\mathbb QB(H)$ whenever $G$ and $H$ are finite groups such that $\dim_{\mathbb Q}\mathbb QB(G)=\dim_{\mathbb Q}\mathbb QB(H)$; see Section 3.3 of [1].
But, if we replace the expression "$G$-set" with "finite $G$-set", the definition of $\mathbb QB(G)$ makes sense for any group $G$, and it seems natural to ask
Question. Do we have $\mathbb QB(G)\simeq\mathbb QB(H)$ if $G$ and $H$ are groups such that $\dim_{\mathbb Q}\mathbb QB(G)=\dim_{\mathbb Q}\mathbb QB(H)$?
The Burnside $\mathbb Q$-algebra $\mathbb QB(G)$ of a group $G$ is canonically isomorphic to that of its profinite completion. The dimension of the Burnside $\mathbb Q$-algebra of a profinite group is finite if and only if the group is finite (but an infinite group may have a finite profinite completion).
For any group $G$ we chose a set $S(G)$ of representatives of the conjugacy classes of finite index subgroups of $G$.
The Burnside $\mathbb Q$-algebra $\mathbb QB(G)$ is von Neumann regular.
Indeed, if $b$ is in $\mathbb QB(G)$, then there is a largest finite index normal subgroup $N$ of $G$ such that $b\in\mathbb QB(G/N)$. Let $\phi_{G/N}:\mathbb QB(G/N)\to\mathbb Q^{S(G/N)}$ be the $\mathbb Q$-algebra isomorphism defined in Section 3.3 of [1], and define $b'\in\mathbb QB(G/N)\subset\mathbb QB(G)$ by $$ b'=(\phi_{G/N})^{-1}\Big(w\circ(\phi_{G/N}(b)\Big), $$ where $w:\mathbb Q\to \mathbb Q$ is defined by $w(\lambda)=\frac1\lambda$ if $\lambda\ne0$ and $w(0)=0$ (that is, $w$ is a witness to the von Neumann regularity of $\mathbb Q$), so that we have $b^2b'=b$ in $\mathbb QB(G)$, which shows that $\mathbb QB(G)$ is von Neumann regular.
(In this post $X\subset Y$ means "$X$ is a (not necessarily proper) subset of $Y$".)
Assume from now on that all the finite index subgroups of $G$ are normal (for instance because $G$ is abelian).
In particular $S(G)$ is the set of all finite index subgroups of $G$.
We will "compute" $\mathbb QB(G)$ in this case.
Using the notation and results in Section 3.2 of [1] we have a $\mathbb Q$-algebra injective morphism $\phi_G:\mathbb QB(G)\to\mathbb Q^{S(G)}$ given by $$ \Big(\phi_G(G/K)\Big)(H)=m(H,K), $$ and we see that $$ \phi_G\left(\frac{G/K}{|G/K|}\right) $$ is the characteristic function $f_K$ of the subset $$ S(G)_{\subset K}:=\{H\in S(G)\ |\ H\subset K\}. $$ Then the $f_K$ with $K\in S(G)$ form a $\mathbb Q$-basis of $\mathbb QB(G)$ satisfying $$ f_Kf_L=f_{K\cap L} $$ for all $K,L\in S(G)$. In particular, up to isomorphism $\mathbb QB(G)$ depends only on the ordered set $S(G)$.
Here are two examples:
$\bullet$ If $k$ is an integer with exactly $n$ prime factors, and if $\mathbb Z_k$ is the group of $k$-adic integers, then the ordered set $S(\mathbb Z_k)$ is opposite to $\mathbb N^n$.
$\bullet$ The ordered set $S(\mathbb Z)$ is opposite to the set of all sequences $x\in\mathbb N^{\mathbb N}$ such that $x_n=0$ for $n$ large enough (depending on $x$).
In general, $\mathbb{Q}B(G)$ is the filtered colimit of $\mathbb{Q}B(H)$ where $H$ ranges over all finite quotients of $G$. In particular, this means $\mathbb{Q}B(G)$ is generated by idempotent elements and is determined up to isomorphism by its Boolean algebra of idempotent elements, which I will call $I(G)$. (See this answer for some related discussion.)
In the case that all finite index subgroups are normal, then $I(G)$ can be identified with the Boolean subalgebra of $\mathcal{P}(S(G))$ generated by the sets $S(G)_{\subseteq K}$ which you consider. This can also be described as the Boolean algebra freely generated by $S(G)$ as a meet-semilattice (proof sketch: reduce to the finite case and use the explicit description of the free Boolean algebra on a finite set).
In particular, if $p$ is prime, then we see that $I(\mathbb{Z}_p)$ is isomorphic to the algebra of finite and cofinite subsets of $\mathbb{N}$, and in particular the ordered set $\mathbb{Z}$ does not embed into $I(\mathbb{Z}_p)$. On the other hand, $\mathbb{Z}$ does embed into $I(\mathbb{Z})$, as exhibited by the following chain of elements of $I(\mathbb{Z})$ (considered as subsets of $S(G)$): $$\dots\subset S(\mathbb{Z})_{\subseteq 8\mathbb{Z}}\subset S(\mathbb{Z})_{\subseteq 4\mathbb{Z}}\subset S(\mathbb{Z})_{\subseteq 2\mathbb{Z}}\subset S(\mathbb{Z})_{\subseteq 2\mathbb{Z}}\cup S(\mathbb{Z})_{\subseteq 3\mathbb{Z}} \subset S(\mathbb{Z})_{\subseteq 2\mathbb{Z}}\cup S(\mathbb{Z})_{\subseteq 3\mathbb{Z}} \cup S(\mathbb{Z})_{\subseteq 5\mathbb{Z}} \subset\cdots$$ Thus $\mathbb{Q}B(\mathbb{Z}_p)$ and $\mathbb{Q}B(\mathbb{Z})$ are not isomorphic, even though they have the same vector space dimension.
(In fact, with a bit more work, you can show $I(\mathbb{Z})$ is atomless, so it must be the free Boolean algebra on countably many generators. Thus $\mathbb{Q}B(\mathbb{Z})$ is isomorphic to the algebra of locally constant $\mathbb{Q}$-valued functions on the Cantor set. Similarly, $\mathbb{Q}B(\mathbb{Z}_p)$ is isomorphic to the algebra of locally constant $\mathbb{Q}$-valued functions on $\mathbb{N}\cup\{\infty\}$, or equivalently the algebra of eventually constant sequences of elements of $\mathbb{Q}$. The latter description can be made very explicit: given a finite $\mathbb{Z}_p$-set, send it to the sequence whose $n$th term is the number of elements which are fixed by $p^n$.)
A natural followup question would be which Boolean algebras are isomorphic to $I(G)$ for some group $G$ (and so which idempotent-generated algebras are isomorphic to $\mathbb{Q}B(G)$ for some group $G$). I don't know how to answer that, but here is a necessary condition: if $I(G)$ is infinite, then it must have a countable descending chain of nonzero elements whose meet is $0$. So, for instance, it is not possible for $I(G)$ to be isomorphic to the Boolean algebra $\mathcal{P}(\mathbb{N})/\mathrm{fin}$.
To prove that such a chain exists, note that any normal finite index subgroup $H\subseteq G$ gives an idempotent $i_H=\frac{G/H}{[G:H]}$, and $K\subset H$ implies $i_K<i_H$ in $I(G)$. If $I(G)$ is infinite, then $\mathbb{Q}B(G)$ is infinite dimensional, so $S(G)$ is infinite. If $H_1,\dots,H_n\subset G$ are finite index subgroups then the kernel of the action on $G/H_1\times\dots\times G/H_n$ is a normal finite index subgroup contained in all of them. So there is no minimal normal finite index subgroup (otherwise it would be contained in all finite index subgroups and so $S(G)$ would be finite), and so we can find an infinite descending chain $G\supset H_1\supset H_2\supset\dots$ of normal finite index subgroups.
I then claim that the meet of the idempotents $i_{H_n}$ is $0$; i.e. no nonzero idempotent of $\mathbb{Q}B(G)$ is less than all of them. To prove this, let $e\in\mathbb{Q}B(G)$ be any idempotent which is less than every $i_{H_n}$ and choose a normal finite index subgroup $H$ such that $e\in\mathbb{Q}B(G/H)$ (i.e., $H$ fixes all the $G$-sets which appear in $e$). Then there is some $n$ such that $H\not\subseteq H_n$, since $[G:H_n]\to\infty$. Now note that no $G$-set appearing in $i_{H_n}e$ is fixed by any element that is not in $H_n$, and so no $G$-set appearing in $i_{H_n}e$ is fixed by all of $H$. But $e\leq i_{H_n}$, so $i_{H_n}e=e$. Since every $G$-set appearing in $e$ is fixed by $H$ but also no $G$-set appearing in $e$ is fixed in $H$, we must have $e=0$.