Which dimensions exist for irreducible quaternionic-type real representations of finite groups?

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.

Let $\mathbf{H}$ be the skew field of real quaternions. Let $Q\subset \mathbf{H}^*$ be the quaternion subgroup of order 8, namely $Q=\{\pm 1,\pm i,\pm j,\pm k\}$.

Let the wreath product $G_n=\mathfrak{S}_n\ltimes Q^n$ (of order $n!8^n$) act on $\mathbf{H}^n$ (viewed as right $\mathbf{H}$-module) as monomial matrices (with nonzero entries in $Q$), namely $Q^n$ acts diagonally: $(q_1,\dots,q_n)\cdot (v_1,\dots v_n)=(q_1v_1,\dots,q_nv_n)$, while $\mathfrak{S}_n$ permutes the coordinates.

Claim: this representation on the real $4n$-dimensional vector space $\mathbf{H}^n$ is irreducible (for $n\ge 1$). Indeed, starting from any nonzero vector $v$, perform a permutation to ensure $v_1=1$, then act by $(-1,0,\dots,0)$ and subtract, to get $v_1\neq 0$ and $v_i=0$ for $i\ge 2$, and finally multiply by elements of $Q$ and use permutations to generate all other vectors.

Then, this can be viewed as an irreducible complex $2n$-dimensional representation, of quaternionic type (and conversely any irreducible complex representation of quaternionic type has positive even dimension).

Actually, in the above argument, $\mathfrak{S}_n$ can be replaced with any transitive subgroup (for the irreducibility argument to hold), and hence we can replace $G_n$ with a subgroup of order $n8^n$.