I'm assuming that the buckets are distinguishable! And the indistinguishable objects are balls.
A hint to answer the question, assuming the balls are indistinguishable
$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})$
In the case of indistinguishability, you will have to use conditioning to get the answer. The calculation remains exactly the same. The only difference is that you are using conditioning instead of combinatorics.
Let $A_i$ be the event that the $i^{th}$ ball chosen falls into bucket 1
$P(A_1\cap A_2 \cap \dots\cap A_8)=P(A_1)\cdot P(A_2|A_1)\cdot P(A_3|A_1\cap A_2)\dots$
There is one way to choose the first ball(since balls are indistinguishable!) and the probability that, that ball goes into the first bucket is $(1/6)$(because the buckets are distinguishable). so, $P(A_1)=1/6$
Now, given that the first ball goes into the first bucket, the number of ways to choose the second ball is also one(balls are indistinguishable!), and the probability that that ball goes into the first bucket is also $(1/6)$(because the buckets are distinguishable). so, $P(A_2|A_1)=1/6$
And so on...
hence $P(\text{all balls fall into bucket 1})=(1/6)^8$ and
$P(X=1)=(6C1)P(\text{all balls fall into bucket 1})=(6C1)(1/6)^8$
I believe, it is straightforward to similarly calculate $P(X=2)$ and $P(X\geq 3)$
I would like to quote @DavidK here(from this answer, you should check out that answer, it is an absolute gem),
A way I think of this intuitively is that we are modeling a world in
which writing a number on a ball or erasing the number does not cause
that ball to magically run away from you when you reach in the back
nor jump into your hand. In fact, the distinguishing marks (or lack
thereof) on the balls have no effect on the probability of drawing a
ball each time. So a correct way to compute $P(X=k)$ with
indistinguishable balls is to compute $P(X=k)$ with distinguishable
balls and simply copy the final result. This yields the same formulas.
I hope, it is okay to quote other users, if it is relevant. If it is not, let me know and I will remove the quote.
Regarding your note:
the probability of dividing 6 objects into exactly 2 buckets is not $[(1/6)^7(5/6) + (1/6)^6(5/6)^2 + (1/6)^5(5/6)^3 + (1/6)^4(5/6)^4 + (1/6)^3(5/6)^5 + (1/6)^2(5/6)^6 + (1/6)(5/6)^7]$
it should be ${nCr}\left(8,1\right)\left(\frac{1}{6}\right)^{7}\left(\frac{1}{6}\right)+{nCr}\left(8,2\right)\left(\frac{1}{6}\right)^{6}\left(\frac{1}{6}\right)^{2}+{nCr}\left(8,3\right)\left(\frac{1}{6}\right)^{5}\left(\frac{1}{6}\right)^{3}+{nCr}\left(8,4\right)\left(\frac{1}{6}\right)^{4}\left(\frac{1}{6}\right)^{4}+{nCr}\left(8,5\right)\left(\frac{1}{6}\right)^{3}\left(\frac{1}{6}\right)^{5}+{nCr}\left(8,6\right)\left(\frac{1}{6}\right)^{2}\left(\frac{1}{6}\right)^{6}+{nCr}\left(8,7\right)\left(\frac{1}{6}\right)^{1}\left(\frac{1}{6}\right)^{7}$