Yes, $G$-structures exist of each finite order. In other words, for every $k\ge1$, there is an $n\ge1$ and a subgroup $G\subset GL(n,\mathbb{R})$ such that its Lie algebra $\frak{g}$ satisfies ${\frak{g}}^{(k-1)}\not=0$ while ${\frak{g}}^{(k)}=0$.

There is no known classification of such algebras, but here is a simple example of an algebra ${\frak{g}}_k\subset {\frak{gl}}(k{+}3,\mathbb{R})$ such that ${\frak{g}}_k$ has order $k$: Let $e_1,\ldots, e_{k+3}$ be the standard basis of $\mathbb{R}^{k+3}$, with dual basis $x^1,\ldots, x^{k+3}$. Let ${\frak{g}}_k$ be the (abelian, nilpotent) subalgebra of ${\frak{gl}}(k{+}3,\mathbb{R})$ with basis $l_1,\ldots,l_k$, where
$$
l_i = e_{i+3}\otimes x^1 + e_{i+2}\otimes x^2.
$$
One computes that ${\frak{g}}_1^{(1)}=0$ and that, for $k>1$, the space ${\frak{g}}_k^{(1)}$ has dimension $k{-}1$, with basis $q_2,\ldots,q_k$, where
$$
q_i = e_{i+3}\otimes (x^1)^2 + 2e_{i+2}\otimes x^1x^2 + e_{i+1}\otimes (x^2)^2.
$$
Continuing on in this way, one finds that the dimension of ${\frak{g}}_k^{(j)}$ is $k{-}j$ for $0\le j\le k$.

For each $n$, there is an upper bound on the order of the subalgebras of ${\frak{gl}}(n,\mathbb{R})$ of finite type, but I do not know what that is. There are estimates for this upper bound, but I don't think they are very tight.

Meanwhile, a theorem of Cartan (originally proved over $\mathbb{C}$ by a classification (but with some omissions), and later completed by others and worked out over $\mathbb{R}$ as well) says that, if $G\subset GL(n,\mathbb{R})$ acts irreducibly on $\mathbb{R}^n$, then $\frak{g}$ has order $1$, $2$, or $\infty$. The list of the irreducibly acting $G\subset GL(n,\mathbb{R})$ that have order $2$ or $\infty$ is known and can be found in my 1996 survey paper, Classical, exceptional, and exotic holonomies: a status report, in Actes de la Table Ronde de Géométrie Différentielle (Luminy, 1992), Sémin. Congr., vol. 1 (1996), pp. 93–165. See the tables in Appendix A.