There are three choices for each of the five balls. Hence, if there were no restrictions, the balls could be placed in the boxes in $3^5$ ways. From these, we must exclude those distributions in which one or more of the boxes is empty.
There are $\binom{3}{1}$ ways to exclude one of the boxes and $2^5$ ways to distribute the balls to the remaining boxes. Hence, there are
$$\binom{3}{1}2^5$$
ways to distribute the balls so that one of the boxes is empty.
However, we have counted those distributions in which two of the boxes are empty twice, once for each of the ways we could have designated one of the empty boxes as the excluded box. We only want to exclude them once, so we must add these cases back.
There are $\binom{3}{2}$ ways to exclude two of the boxes and one way to place all the balls in the remaining box.
Hence, the number of ways the balls can be distributed so that no box is left empty is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
by the Inclusion-Exclusion Principle.
Where am I going wrong?
You count each distribution in which one box receives three balls and the others receive one three times, once for each way you could place one of those three balls first.
You count each distribution in which two of the boxes receive two balls and the other box receives one four times, once for each way you could place one of the two balls in each of the two boxes with two balls first.
Three balls in one box and one ball in each of the others: There are three ways to choose which box receives three balls, $\binom{5}{3}$ ways to choose which three balls are placed in that box, and $2!$ ways to distribute the remaining balls. Hence, there are
$$\binom{3}{1}\binom{5}{3}2!$$
ways to distribute the balls so that three balls are placed in the same box.
Two boxes receives two balls and one box receives one ball: There are three ways to choose which box receives only one ball and five ways to choose the ball that is placed in that box. There are $\binom{4}{2}$ ways to choose which two of the remaining four balls are placed in the smaller of the two remaining boxes. The other two balls must be placed in the remaining box. Hence, there are
$$\binom{3}{1}\binom{5}{1}\binom{4}{2}$$
ways to distribute the balls so that two boxes receive two balls and one box receives one.
Observe that
$$\binom{3}{1}\binom{5}{3}2! + \binom{3}{1}\binom{5}{1}\binom{4}{2} = 3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5$$
Since you counted distributions in which one box receives three balls and the others receive one three times and distributions in which two boxes receive two balls and the other receives one four times, you obtained
$$3\binom{3}{1}\binom{5}{3}2! + 4\binom{3}{1}\binom{5}{1}\binom{4}{2} = \binom{5}{3} \cdot 3! \cdot 3^2$$
First assume that the boxes are distinct. Then the number of ways of distributing 7 distinct balls into these boxes so that none is empty is, by Principle of Inclusion Exclusion,
$$4^7 - \binom{4}{1}3^7 + \binom{4}{2}2^7 - \binom{4}{3} 1^7 = 8400$$
Now the naming of the boxes can be done in $4! = 24$ ways, the required number is $\dfrac{8400}{24} = 350$.
In fact if we want to distribute $n$ distinct objects into $k$ identical boxes so that no box is empty, the number of ways is given by the Stirling number $S(n,k)$ and
$$S(n,k) = \frac{1}{k!}\sum_{i=0}^{k-1} \binom{k}{i}(k-i)^n$$
Best Answer
Your first method is correct.
Your second method does indeed count each distribution multiple times.
The possible distributions are $(4, 1, 1)$, $(3, 2, 1)$, $(3, 1, 2)$, and $(2, 2, 2)$.
There are four distributions of the form $(4, 1, 1)$, depending on whether the left or right square is left empty in the second and third rows. Your second method counts such distributions four times, once for each way you could have designated one of the four objects in the first row as the object in the first row.
There are eight distributions of the form $(3, 2, 1)$, depending on which of the four squares is left empty in the first row and which of the two squares is left empty in the third row. Your second method counts such distributions $3 \cdot 2 = 6$ times, once for each way you could have designated one of the objects in the first row as the object in the first row and once for each way you could have designated one of the objects in the second row as the object in the second row.
By symmetry, there are eight distributions of the form $(3, 1, 2)$, each of which you have counted six times.
There are six distributions of the form $(2, 2, 2)$, depending on which two entries have been left empty in the first row. Your second method counts each such distribution $2 \cdot 2 \cdot 2 = 8$ times, once for each way you could have designated one of the two objects in each row as the object in that row.
Notice that $$\color{red}{4} \cdot 4 + 2 \cdot \color{red}{6} \cdot 8 + \color{red}{8} \cdot 6 = \color{red}{16} + \color{red}{96} + \color{red}{48} = \color{red}{160}$$