Now that I have thought about the question some more, I can give a better answer. I have a remark about how to search for these resolutions in general, and a construction that leads to many examples.
First, every permutation module is part of many exact sequences of permutation modules in which the subgroup is trivial: $$\cdots \to R[G]^{n_2} \to R[G]^{n_1} \to R[G/H] \to 0.$$ The reason is very simple and standard: Any exact complex of this type is by definition a free resolution. The way that you make a free resolution is that there is some intermediate target or kernel $K$, and you can send the generators $1 \in R[G]$ to some spanning set of $K$. Usually the resolution is infinite, but with standard linear algebra you can search for a finite solution when there is one.
This observation generalizes to other permutation modules. There is a part of induction-restriction reciprocity that holds over any ring. Namely, $$\text{Hom}_G(R[G/H],M) \cong \text{Hom}_H(I,M),$$ where $I$ is the trivial representation. This relation is a generalization of the proof that a free module is projective. So if there is a finite permutation resolution of a module $M$ (which could be the kernel of some incomplete sequence of permutation modules), you can search for it in the same way that you search for free resolutions.
Second (and I suspect that readers will like this answer better), you can obtain many examples from the chain homology complex of a finite CW complex $K$ with an action of $G$. In order to make everything match, let's consider a slight generalization of a permutation module, not just $R[G/H]$ but also a module $R[G/H]_\chi$ induced from a character $$\chi:H \to \{1,-1\}.$$ The basic idea is not hard: Each term $C_n(K)$ is a direct sum of signed permutation modules of $G$, because $G$ acts on the cells. If a cell $c$ has stabilizer $H$, then it makes an orbit equivalent to $G/H$, and we can define $\chi$ by examining which elements of $H$ flip over $c$. If $K$ happens to have the same $R$-homology as a point, then you can augment its chain complex by the trivial module. Or if it is an $R$-homology sphere, you can augment its chain complex at both ends. If you don't like the signed permutation modules, you can subdivide the cell $c$ to get rid of them, or work in characteristic 2.
If $K$ is a line segment and $G = C_2$ acts by reflecting it, the result is Leonid's first example.
If $K$ is a polygon with $n$ sides and $G = C_n$ acts by rotation, the result is Leonid's second example.
If $K$ is a polygonal tiling of the 2-sphere and $G$ is a rotation group that acts on $K$ without reversing edges, the result is a new example. For instance you can take a dodecahedron graph and divide each edge into two edges. Again, the point of splitting the edges is just to get rid of the signed permutation modules.
Every finite group $G$ acts faithfully on a sphere of some dimension, because $G$ has a faithful linear representation. So there are many sphere examples for every finite group.
At first glance, the second answer is a type of construction. In many cases, it is also an interpretation of a chain complex $C$, because if you have $C$ you can try to build a CW complex to represent it. In order to be an augmented chain complex, $C$ needs to end in the trivial module $I$. If it ends in something else, you can concatenate with $$0 \to I \to I \to 0.$$ The differentials of $C$ also need to have integer matrices.