You made your mistake in counting the number of cases that violate the restriction that the lower four drawers can contain at most three balls.
Let $x_i$ be the number of balls placed in drawer $i$, where we count from the top drawer down. Since eight balls are placed in the five drawers,
$$x_1 + x_2 + x_3 + x_4 + x_5 = 8 \tag{1}$$
is an equation in the nonnegative integers. The restriction that at most four balls may be placed in the top drawer means $x_1 \leq 4$. The restriction that at most three balls may be placed in every other drawer means $x_2, x_3, x_4, x_5 \leq 3$.
A particular solution of equation 1 corresponds to the placement of $5 - 1 = 4$ addition signs in a row of eight ones. For instance,
$$1 1 + 1 + 1 1 1 + + 1 1$$
corresponds to the solution $x_1 = 2$, $x_2 = 1$, $x_3 = 3$, $x_4 = 0$, $x_5 = 2$. The number of such solutions is the number of ways we can place four addition signs in a row of eight ones which is
$$\binom{8 + 5 - 1}{5 - 1} = \binom{12}{4}$$
since we must choose which four of the twelve positions required for eight ones and four addition signs will be filled with addition signs.
From these, we must count those cases that violate one or more of the restrictions. Let $A_i$ be the set of solutions in which the $i$th drawer has more than the permitted number of balls.
$|A_1|$: The top drawer is allowed to have at most four balls, so we must subtract those cases in which $x_1 \geq 5$. Suppose $x_1 \geq 5$. Then $x_1' = x_1 - 5$ is a nonnegative integer. Substituting $x_1' + 5$ for $x_1$ in equation 1 yields
\begin{align*}
x_1' + 5 + x_2 + x_3 + x_4 + x_5 & = 8\\
x_1' + x_2 + x_3 + x_4 + x_5 & = 3 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{3 + 5 - 1}{5 - 1} = \binom{7}{4}$$
solutions.
$|A_2|$: The second drawer is allowed to have at most three balls, so we must subtract those cases in which $x_2 \geq 4$. then $x_2' = x_2 - 4$ is a nonnegative integer. Substituting $x_2' + 4$ for $x_2$ in equation 1 yields
\begin{align*}
x_1 + x_2' + 4 + x_3 + x_4 + x_5 & = 8\\
x_1 + x_2' + x_3 + x_4 + x_5 & = 4 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers with
$$\binom{4 + 5 - 1}{5 - 1} = \binom{8}{4}$$
solutions.
By symmetry,
$$|A_2| = |A_3| = |A_4| = |A_5| = \binom{8}{4}$$
$|A_1 \cap A_2|$: It is not possible for both the restrictions $x_1 \leq 4$ and $x_2 \leq 3$ to be violated since we would require at least $5 + 4 = 9 > 8$ balls to be placed in the drawers.
By symmetry,
$$|A_1 \cap A_2| = |A_1 \cap A_3| = |A_1 \cap A_4| = |A_1 \cap A_5| = 0$$
$|A_2 \cap A_3|$: The only way to violate both the restrictions $x_2 \leq 3$ and $x_3 \leq 3$ is to place four balls in each drawer, which can be done in one way.
By symmetry,
$$|A_2 \cap A_3| = |A_2 \cap A_4| = |A_2 \cap A_5| = |A_3 \cap A_4| = |A_3 \cap A_5| = |A_4 \cap A_5| = 1$$
It is not possible to violate three or more restrictions at once since we only have eight balls and $3 \cdot 4 = 12 > 8$.
Hence, by the Inclusion-Exclusion Principle, the number of admissible ways to distribute the balls is
$$\binom{12}{4} - \binom{7}{4} - \binom{4}{1}\binom{8}{4} + \binom{6}{1}\binom{4}{4} = 495 - 35 - 4 \cdot 70 + 6 \cdot 1 = 495 - 35 - 280 + 6 = 186$$
Best Answer
Here's maybe a more helpful way to think through how these numbers are obtained.
To compute $A$, imagine the American block is just another person to be arranged among the $8$ others. Then we have $9$ "people," giving $9!$ arrangements. We must then multiply by $4!$ to arrange the Americans within the block.
For $A \cap R$, we have $7$ "people" (5 actual people and 2 blocks), giving $7!$, multiplied by $4! \cdot 3!$ to arrange within the blocks.
Similarly, for $A \cap R \cap C$, we must arrange the 3 blocks ($3!$) and then arrange within the blocks ($4! \cdot 3! \cdot 5!$)