Any suggestions on where I made an error in my logic?
The problem with your logic is that you are introducing an extra restriction. You are allowing only the last ball to be paired with another ball. In other words you are forcing the last ball to necessarily be in the bin with $2$ balls. This is a restriction that the question doesn't have.
Here's how we can correct it. What you are doing with the last ball can be done with any of the $n$ balls. So multiply by $n$.
However, putting the $i^{th}$ ball in the bin containing the $j^{th}$ ball is the same as putting the $j^{th}$ ball in the bin containing the $i^{th}$ ball. So we need to divide by $2$.
So the correct answer would be
$$\frac{(n-1)(n-1)!}{n^{n-1}} \times \frac{n}{2} = \frac{(n-1)(n-1)!}{2 n^{n-2}}$$
Response to OP's comment
Let me rephrase your solution from the start. Perhaps it will help you gain an intuitive understanding.
Let's find the probability that $i^{\text{th}}$ ball ends up with the $j^{\text{th}}$ ball (as a pair) in one bin and the other balls go into separate bins. We will label this $P(E_{ij})$
For this, imagine placing each ball into a separate bin one by one, with one caveat. When we get to the $j^{\text{th}}$ ball, we hold on to it and move on to the next ball. at the end we will have placed $n-1$ balls in $n-1$ bins and we will have the $j^{\text{th}}$ ball with us. The probability for this is
$$\frac{n}{n} \cdot \frac{n-1}{n} \cdot \frac{n-2}{n} \cdots \cdot \frac{2}{n} = \frac{n!}{n^{n-1}}$$
Now, the ball still with us ($j^{\text{th}}$) must go into one specific bin - that with the $i^{\text{th}}$ ball. The probability for this happening is $\dfrac{1}{n}$. Multiplying this with the above expression, we get
$$P(E_{ij}) = \frac{n!}{n^{n-1}} \times \frac{1}{n} = \frac{(n-1)!}{n^{n-1}}$$
As you can see $P(E_{ij})$ is a constant independent of $i$ and $j$. Now, to get the total probability for exactly one bin to be empty we need to add it for all pairs of balls. Or multiply it with the number of pairs possible, $\dfrac{n (n-1)}{2}$. So
$$\begin{align*}
P &= \frac{(n-1)!}{n^{n-1}} \times \frac{n (n-1)}{2} \\[0.3cm]
&= \frac{(n-1)(n-1)!}{2n^{n-2}}
\end{align*}$$
Best Answer
The answer is assuming that the balls can only land in the three buckets and each bucket is equally likely as a target.
The easier way to do it would be thinking go it in the other direction: What is the probability that at least one bucket has no balls?
Let the aforementioned probability be: P
Therefore, the answer you require is: 1 - P.
To calculate P:
$P = $ probability that ball lands only in any 2 buckets($=P_1$) $+$ probability that ball lands in only one bucket($=P_2$) $$ P_1 = {3\choose 2} (\frac{2}{3})^{10} $$ as, you choose any two buckets in which are the balls are supposed to fall ${3\choose 2}$ and the probability that the ball falls only in those two is $\frac{2}{3}$.
Since you are throwing 10 balls, simply multiple $\frac{2}{3}$ 10 times to give $P_1$ $$ P_2 = {3\choose 1} (\frac{1}{3})^{10} $$ as, you choose only buckets in which are the balls are supposed to fall ${3\choose 1}$ and the probability that the ball falls only in that bucket is $\frac{1}{3}$.
Since you are throwing 10 balls, simply multiple $\frac{1}{3}$ 10 times to give $P_1$
Therefore, your answer is: $$ 1 - P_1 - P_2 = 1 - {3\choose 2} (\frac{2}{3})^{10} - {3\choose 1} (\frac{1}{3})^{10} $$