Observe that since the items are identical, it does not matter that there are $4$ items in the second box. Your are then asking for the number of sums $a_1+a_2+a_3=3$ where $0\leq a_1\leq 1$, $0\leq a_2\leq 3$, and $0\leq a_3\leq 2$. I will give three answers.
First, an elementary argument: We know that $a_1=0$ or $a_1=1$. If $a_1=0$, then $a_2+a_3=3$. In this case, there are three possibilities: $3+0=3$, $2+1=3$, and $1+2=3$. If $a_1=1$, then $a_2+a_3=2$ and there are still three possibilities: $2+0=2$, $1+1=2$, and $0+2=2$. This results in $6$ different options.
A more combinatorial argument: The number of ways to write $n$ as a sum of $k$ nonnegative integers is
$$
\binom{n+k-1}{k-1}
$$
and a discussion can be found here.
So, in this case, the number of ways that $a_1+a_2+a_3=3$ (without restrictions) is
$$
\binom{3+3-1}{3-1}=\binom{5}{2}=10.
$$
This, however, counts too many possible sums. Suppose that we take too many from box $1$, this means that we take at least $2$ from box $1$. In this case, we can write $a_1=2+b_1$ where $b_1$ is nonnegative. Then, the initial sum becomes $b_1+a_2+a_3=1$. Using the same formula, this results in
$$
\binom{1+3-1}{3-1}=\binom{3}{2}=3
$$
impossible ways. Continuing, there is no way to take too many objects from the second box, but it is possible to take too many objects from box $3$. In this case, one must take $3$ objects from box $3$, so we write $a_3=3+b_3$ where $b_3$ is nonnegative. This results in the equation $a_1+a_2+(3+b_3)=3$. There are
$$
\binom{0+3-1}{3-1}=1
$$
ways for this sum to occur. We should now use the inclusion/exclusion principle to see if we've over-counted. This could happen if we take more than $1$ item from box $1$ and more than $2$ items from box $3$. Then, we have $(2+b_1)+b_2+(3+b_3)=3$, but this has no solutions as a sum of nonnegative integers cannot be negative.
Therefore, out of the original $10$ possibilities, $3+1=4$ are impossible, leaving the $6$ that we've found.
A dynamic programming-type solution: Let $N(b,s)$ be the number of ways to use the first $b$ boxes to sum to $s$. Also, write $m_i$ for the number of objects in box $i$. In your case:
\begin{align*}
N(1,0)&=1\\
N(1,1)&=1\\
N(1,2)&=0\\
N(1,3)&=0.
\end{align*}
Then, the values in the second box can be computed as follows:
$$
N(b+1,s)=\sum_{i=0}^{\min\{s,m_{b+1}\}}N(b,s-i).
$$
Using this formula:
\begin{align*}
N(2,0)&=N(1,0)=1\\
N(2,1)&=N(1,0)+N(1,1)=2\\
N(2,2)&=N(1,0)+N(1,1)+N(1,2)=2\\
N(2,3)&=N(1,0)+N(1,1)+N(1,2)+N(1,3)=2.
\end{align*}
Continuing for the third column,
\begin{align*}
N(3,0)&=N(2,0)=1\\
N(3,1)&=N(2,0)+N(2,1)=3\\
N(3,2)&=N(2,0)+N(2,1)+N(2,2)=5\\
N(3,3)&=N(2,1)+N(2,2)+N(2,3)=6.
\end{align*}
We are interested in the value $N(3,3)=6$.
