HINT: Representing the six colors by the numbers $1$ through $6$, we see that the following arrangements are equivalent:
$$\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\qquad\begin{bmatrix}3&2&1\\6&5&4\end{bmatrix}\qquad\begin{bmatrix}4&5&6\\1&2&3\end{bmatrix}\qquad\begin{bmatrix}6&5&4\\3&2&1\end{bmatrix}$$
The first is the original; the second is flipped about the vertical axis; the third is flipped about the horizontal axis; and the last is rotated $180^\circ$ degrees (or flipped about each of the axes).
What should you divide by?
We can place a surprisingly tight bound on the answer using two simple models. Here the models are explored for the simpler case of a tetrahedron withbone type of dot, then the results are applied to the icosahedron.
Model 1: Many Colors
Since the different orientations of one or two dots are considered not equivalent, we suppose that there are not four but eight different colors possible for each face. We then apply the Polya theory to eight colors on a tetrahedron to get
$N_1(tet,1)=(1/12)[8^4+(11×8^2)]=400$
The problem with this us that not all colors are fully distinct. As described in a comment to my original answer, this may double-count some combinations due to "colors" rotating into one another which actual colors don't do. So we should treat this result as an upper bound.
Model 2: Limits of Symmetry
If all the dot positions were truly distinct, we would simply count off $2^{12}$ possible choices of dot or no dot on the tetrahedron. That would give $4096$. But what if some of these orientations are rotatable into each other? In this model we would not really know. But we can observe that no more than twelve configurations at a time can be rotated intobeach other because that is all the rotational symmetry therebus in a tetrahedron! So we are sure of the lower bound
$N_2=4096/12=342$ (rounded up, can't have a fraction)
We are surprised at how tight these bounds are. We could surely call the true count $370$ and bet the house that we are within ten percent for the one-type tetrahedral case!
And the relative error only gets better with the icosahedron ... .
Icosahrdron, one type of dot
With $2^3=8$ eight colors on each face for Model 1 and sixty dots for Model 2, we render the bounds thusly:
Upper Bound (Model 1): $N_1(icosa,1)=(1/60)[8^{20}+(24×8^4)+(15×8^{10})+(20×8^8)]\approx1.922×10^{16}$
Lower Bound (Model 2):: $N_2=2^{60}/60\approx1.922×10^{16}$(!)
We can report that to four siginficant digits, the permutation count is $1.922×10^{16}$! (The exact figure was later calculated by a more sophisticated use of the Polya equation: $1921538678935736$.)
Icosahrdron, two types of dot
With three choices at each position of each face Model 1 gives $3^3=27$ colors. Model $2$ gives $3^{60}$ dot permutations before dividing out rotational symmetries:
Upper Bound (Model 1): $N_1(icosa,2)=(1/60)[27^{20}+(24×27^4)+(15×27^{10})+(20×27^8)]\approx7.065×10^{26}$
Lower Bound (Model 2):: $N_2=3^{60}/60\approx7.065×10^{26}$(!)
Again the simple models bracket (at least) four significant digits!
Figuring out the trick
Like any good magician, I subtly apllied some misdirection. Let's go back to the tetrahedral, one-type case. In Model 1 I apply the Polya theory, with the first term divided by the rotational symmetry ($12$) being $8^4$. In the second case I identify just a single term divided by $12$, namely $2^{12}$. Well ... these two quantities are the same, and moreover that matched term dominates $N_1$ because the other tetms have smaller exponents on the base $8$. So we get a tight bound because the dominant terms agree. Going to the icosahedron only tightens this agreement because with more dots to play with, the ecponents and the differences between said exponents goes up even higher.
Best Answer
You can view an arrangement as an element of a set $S$. The total number of elements in $S$ is $8!$ (without any restrictions). Consider any element, for example; $$ \begin{bmatrix} 1 & 2 & 3 & 4\\ 5 & 6 & 7 & 8 \end{bmatrix} $$ (each number corresponds to a color). Now, we can get $3$ additional elements of $S$ just by a $180^\circ$ rotation, a vertical flip or a horizontal flip, giving you $$ \begin{bmatrix} 8 & 7 & 6 & 5\\ 4 & 3 & 2 & 1 \end{bmatrix}, \begin{bmatrix} 4 & 3 & 2 & 1\\ 8 & 7 & 6 & 5 \end{bmatrix}, \begin{bmatrix} 5 & 6 & 7 & 8\\ 1 & 2 & 3 & 4 \end{bmatrix} $$
Notice that applying these operations to any one of $4$ elements results in an element from one of these only. Hence, you can partition $S$ into such independent groups of $4$. Therefore the total number of arrangements identical upto rotations and flips is $\dfrac{8!}{4}$.
Problem in your solution:
You can now see why the solution $\dfrac{8!}{8}$ is not correct. When you first divide by $2$, let's say to remove $180^\circ$ rotations, half of the elements are rightly removed. The next time you divide by $2$, the horizontal flips are removed. To remove the vertical flips, you've divided by $2$ once more, but notice that a $180^\circ$ rotation followed by horizontal flip is essentially a vertical flip, so by just dividing by $4$ you've removed even the vertically flipped versions of every element in the process, hence dividing by $2$ once more leads to a miscount.