Take a solved cube and do $RU$; then trace the cycle structure of the permutation it realizes.
The combination moves 5 corner cubies in a cycle where a cubie is twisted by a third of a turn when it gets back to its original position, so that's a factor of 15.
It also twists a the FRU corner by one third of a turn; that's taken care of by the factor of 15 too.
Then it permutes 7 edges cyclically, but this time every edge has the correct orientation when it gets back.
So the order is the least common multiple of 7 and 15, namely 105.
(For a subercube we need another factor of 4 to get the centers back into the original orientation).
The maximum orders for a $n\times n\times n$ Rubik's cube group:
\begin{align*}n=2: &\quad 3^2\cdot 5 = 45\\
n=3: &\quad 2^2\cdot 3^2\cdot 5\cdot 7 = 1260\\
n=4: &\quad 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 765765\\
n=5: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520\\
n\ge 6: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880\end{align*}
In general, there are five types of piece to consider:
- For odd $n$, each face has a center. By convention, we treat these as fixed. If we don't, they can be rotated like a solid cube, for a copy of $S_4$. Highest order $4$, least common multiple $2^2\cdot 3=12$. As the largest possible order is divisible by $12$ for all odd $n\ge 3$, whether we include these doesn't affect the answer.
- For any $n\ge 2$, there are $8$ corner pieces which can be permuted and rotated, a semidirect product of $S_8$ and $(\mathbb{Z}/3)^8$ - except that that the "sum" of those rotations is zero, dropping one factor of $3$ from the latter part. The order of an element is at worst $3$ times the order of an element of $S_8$. Highest possible order $15\cdot 3=45$, least common multiple $2^3\cdot 3^2\cdot 5\cdot 7$.
- For any odd $n\ge 3$, there are $12$ centers of edges, which can be permuted and reflected. The number of reflections must also be even, for a semidirect product of $S_{12}$ and $(\mathbb{Z}/2)^{11}$. Highest possible order $60\cdot 2=120$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11$.
- For any $n\ge 4$, there are $\left\lfloor\frac{n}{2}\right\rfloor-1$ orbits of 24 off-center edges, which can't change how far from the center of their edge they are. The pair of edges on each side are distinguishable due to being reflections of each other, and the group for each set of these pieces is a copy of $S_{24}$. Highest possible order $8\cdot 7\cdot 5\cdot 3 = 840$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$.
- For any $n\ge 4$, there are $\left\lfloor\frac{(n-2)^2}{4}\right\rfloor$ orbits of 24 off-center face pieces each. These can be rotated in their face, but not reflected; mirror images across a symmetry line within a face (in size $\ge 6$) are separate orbits. Each piece of the same color in one of these sets is indistinguishable, so that gets us an action of $S_{24}$, with $\frac{24!}{24^6}$ possible states. The possible orders are the same as in the previous case. In terms of what we need to solve the cube, there are three variants based on whether they lie on a symmetry line of a face - but they're all the same for the count we care about here.
So then, the highest possible order in any Rubik group is the least common multiple of all of these, namely $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$. This is achieved for all $n\ge 6$; we have at least six of the size-24 orbits, which we can use to produce orders of $16\cdot 5$, $9\cdot 11$, $7\cdot 13$, $17$, $19$, and $23$.
Now, all of this assumes that we can modify the pieces in each orbit independently. That isn't quite true - several combinations must be even permutations. First, for odd $n$, the combined permutation of corners and edge centers is even; they're modified by face rotations, each of which is a 4-cycle of edges and a 4-cycle of corners. Second, for $n\ge 4$, the combined permutation of the corners and an orbit of pieces on face diagonals is even. Next, for $n\ge 6$, each mirrored pair of orbits of off-center face pieces must have a combined even permutation; each face rotation and each relevant slice rotation is a 4-cycle in each of the orbits. Finally, for odd $n\ge 5$, the combined permutation of off-center edge pieces, pieces on face diagonals, and pieces on face midlines at a fixed distance from the center is even. These parity restrictions are the only additional restriction; other than that, the orbits can be manipulated independently.
How does this change things? For the $n\ge 6$ case, all we have to do is implement the order $16\cdot 5$ piece as the product of a $16$-cycle, a $2$-cycle, and a $5$-cycle. Then all of the permutations we're working with are even, so there aren't any problems. In the smaller cases, we'll have to check.
For $n=2$, only the corners matter, and the highest order is $45$.
For $n=3$, we have corners and edge centers. With $17$ possible orders for the first and $31$ for the second, we can just scan all of the possible combinations. The highest LCM is $2520$, from order $45$ (3-cycle, 5-cycle, rotations in the 3-cycle don't add to zero) in the corners and order $56$ (4-cycle, 7-cycle, reflections in the 4-cycle don't add to zero) in the edges. This, however, doesn't pass the parity check; the combined permutation is odd, and there's no wiggle room to change that. That leaves us dropping down to the second-highest possibility of $1260$, possible in several ways ($45$ and $28$, $45$ and $84$, $35$ and $36$, $70$ and $36$). Three of those fail the parity check, leaving only the pair of order $45$ in the corners and order $28$ (Two 2-cycles, a 7-cycle, reflections in at least one 2-cycle don't add to zero) in the edges. Highest order $1260$, as previously noted.
For $n=4$, we have the corners, one orbit of off-center edges, and one orbit of off-center face pieces. With over a hundred possible orders in $S_{24}$, finding the least common multiple isn't the easiest search. Pruning everything with an order that divides another gets that down to a manageable total of $27$, and $5$ for the corners. Running it in the spreadsheet I've been using, the LCM is $765765=3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$; $45$ in the corners, $11\cdot 13$ in the edges, and $7\cdot 17$ in the faces. Checking parity... all the pieces are odd permutations, so we'll actually have something of that order.
For $n=5$, we have corners, edge centers, three size-24 orbits, and two parity restrictions that cross orbits. The best I've got here is $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520$; it could use an independent check, but I'm pretty sure it's right. The three size-24 orbits get us orders of $23$, $7\cdot 17$, and $11\cdot 13$ while the corners give us order $45$ and the edges give order $16$ (an 8-cycle and a 2-cycle, with an odd number of reflections in the 8-cycle). They're all odd permutations, so the parity checks pass.
For $n\ge 6$, as previously noted, we reach the maximum possible order $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880$.
Best Answer
Most explanations of the order of the cube group don't use a group-theoretic wording because it is felt to be easier to understand it in more concrete terms, given that the cube is a physical object.
But if you want to, we can certainly phrase the same calculation in more abstract group-theoretic language.
In that case, we could start by defining $G$ to be a certain subgroup of the group $S_{54}$ of permutations of all the color stickers, given by its six generators.
$G$ acts on the 20 movable cubies, so there's a homomorphism $G\to S_{20}$. Its kernel is a group of $H$ operations that twist and flip cubies in-place, and the quotient is the group of legal cubie permutations. To count the order $G$ itself we multiply the order of the kernel with the order of the quotient.
The quotient, as a subgroup of $S_{20}$ turns out to be $A_{20}\cap(S_{12}\times S_{8})$, which we can prove by noting that each of the 6 generators is in this subgroup, and finding a set of generators for $A_{20}\cap(S_{12}\times S_{8})$ that can be realized as concrete cube operations.
The kernel $H$ (the group of twists-and-flips that don't permute the cubies) is a subgroup of $G$, but is is clearly abelian and isomorphic to a subgroup of $(C_2)^{12}\oplus(C_3)^8$, and I will view it as such henceforth. $H$ contains each of its projections to $(C_2)^{12}$ and $(C_3)^{8}$ -- namely, they are $\{x^3\mid x\in H\}$ and $\{x^4\mid x\in H\}$, respectively (this can be generalized to any subgroup of a direct sum of abelian groups of coprime orders) -- and is therefore their direct sum, and its order is the product of the orders of the components.
Consider the intersection with the corner-twist group $(C_3)^8$. We can count it by viewing it as a subspace of $(\mathbb F_3)^8$ and applying linear algebra (and it is a subspace: multiplying by an scalar from $\mathbb F_3$ just corresponds to iterating the operation). It is easy to show that it has dimension $\ge 7$, by showing 7 linearly independent elements, say, combinations that turn the FRU corner together with each of the 7 other ones. So if only we can show that it has dimension $<8$, it will follow that it and has $3^7$ elements.
To show this, assign a canonical orientation of each cubie in each location as described in this answer. Each configuration of the full cube now describes an element of $(\mathbb F_3)^8$, with components given by how the cubie that happens to be found in a given position is twisted from its canonical orientation. The entire $G$ then acts on $(\mathbb F_3)^8$, and each action preserves the hyperplane $x_1+x_2+\cdots+x_8=0$ (which can be checked for each of the six generators). The corner-twist subspace is contained in the orbit of $(0,0,\ldots,0)$, which is contained in the hyperplane, and therefore cannot be all of $(\mathbb F_3)^8$.
The handling of the edge-flip group is the same, mutatis mutandis.
Is all of this more enlightening than the usual more concrete way of counting? I doubt it.