One source of examples in dimension $2^{n}$ for any positive integer $n$ is given by extraspecial $2$-groups of order $2^{2n+1}.$
I won't give all details, but one can argue as follows: if $S$ is an extraspecial $2$-group of order $2^{2n+1},$ then the number of elements not of order $4$ in $S$ is given by $2^{n}(2^{n}+ \nu(\chi)),$ where $\nu(\chi)$ is the Frobenius-Schur (henceforth F-S)indicator the unique non-linear complex irreducible character $\chi$ of $S$. This uses the well-known formula that the number of solutions of $x^{2} = 1$ in $S$ is given by $\sum_{\mu \in {\rm Irr}(S)} \nu(\mu) \mu(1).$
For any positive $n,$ there are two non-isomorphic extraspecial groups of order $2^{2n+1},$ and they are distinguishable because they have different numbers of elements of order $4$. Each of those groups have $2^{2n}$ linear characters (all realizable over $\mathbb{R}),$ and one real-valued irreducible character of degree $2^{n}.$
Hence one of the groups has the real-valued irreducible character of degree $2^{n}$ with F-S indicator $1$ and one has such a character with F-S indicator $-1$. The latter comes from a quaternionic representation.
If you want to know which extraspecial group has the quaternionic representation, you can argue as follows.
One of the extraspecial groups is (up to isomorphism) a central product of $n$ copies of $D_{8}$, and the other is (up to isomorphism) a central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}.$ The first of these groups clearly has a representation of degree $2^{n}$ which is realizable over $\mathbb{R}$ as $D_{8}$ has a $2$-dimensional absolutely irreducible representation over $\mathbb{R}.$
Hence the $2^{n}$-dimensional faithful representation of a central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}$ is a quaternionic representation.
Later edit: Note that (as is well-known) the central product of $n$ copies of $D_{8}$ therefore has $2^{2n}-2^{n}$ elements of order $4$ and that the central product of $n-1$ copies of $D_{8}$ with one copy of $Q_{8}$ has $2^{2n} +2^{n}$ elements of order $4$.
Later edit in a different (more general) direction: If you have enough information about the irreducible characters of a finite group $G,$ then it is easy to check whether $G$ has quaternionic representations or not. If the sum of the degrees of the non-trivial real-valued complex irreducible characters of $G$ is equal to the number of involutions of $G,$ then $G$ has no quaternionic irreducible representations, and in any other case, $G$ does have (necessarily non-trivial) irreducible quaternionic representations.
In the latter case, when $G$ is also simple, any quaternionic representation is likely to have relatively large degree, since for any integer $m >1$, there are only finitely many simple groups with an $m$-dimensional complex irreducible representation of degree $m$.