There are no examples with $R$ local and $M$ finitely generated. I think you should be able to reduce the general case to the local case, and you can certainly do so when $R$ is noetherian, but I have a nagging suspicion that the property "torsion-free" does not localize as well as I want it to for non-noetherian rings. I don't have much intuition for the non-finitely-generated case.
$\def\mm{\mathfrak{m}}$ Proof: Suppose that $M$ is not flat. Let $R$ be local, $\mm$ the maximal ideal and $k = R/\mm$. Let $V = M \otimes \mm$. Let $f_i$ be a basis of $V$, for $i=1$ to $n$, and let $e_i$ in $M$ be a preimage of $f_i$. By Nakayama's lemma, the map $R^{\oplus n} \to M$ sending $(x_1, \ldots, x_n)$ to $\sum x_i e_i$ is surjective so, if $M$ is not flat, it must have a kernel. In other words, there must be some $(x_1, \ldots, x_n)$ in $\mm^n$, not all $0$, such that $\sum x_i e_i=0$. Without loss of generality, let $x_n$ be nonzero.
Set $$\Delta := \sum_{\sigma \in S_n} \epsilon(\sigma) e_{\sigma(1)} \otimes \cdots \otimes e_{\sigma(n)} \in M^{\otimes n}.$$ Here $\epsilon(\sigma)$ is the sign of the permutation $\sigma$. I claim that $\Delta$ is nonzero but $x_n \Delta =0$.
Proof that $\Delta$ is nonzero: By the associativity of tensor product, $M^{\otimes n} \otimes k \cong V^{\otimes n}$. The image of $\Delta$ in $V^{\otimes n}$ is nonzero, so $\Delta$ is nonzero.
Proof that $x_n \Delta=0$ is zero: Note that $$x_n e_1 \otimes \cdots \otimes e_n = e_1 \otimes \cdots \otimes e_{n-1} \otimes \left(- x_1 e_1 -x_2 e_2 -\cdots - x_{n-1} e_{n-1} \right).$$ Similarly expand each of the $n!$ terms. You get an antisymmetric expression of degree $n$ in $e_1$, ..., $e_{n-1}$, so it must be zero.