To solve this problem, we can start by considering the given information.
Let's assume the length, width, and height of the rectangular box are $l$, $w$, and $h$ respectively.
Given that the volume of the box is $4320$ cubic inches, we have the equation:
$$lwh = 4320 \qquad \text{(Equation 1)}$$
It is also given that the surface area of the box is $1704$ square inches. The surface area of a rectangular box can be calculated using the formula:
$$2lw + 2lh + 2wh = 1704 \qquad \text{(Equation 2)}$$
Lastly, we are told that the sum of the lengths of all $12$ edges is $208$ inches. Each edge appears twice when calculating the sum, so we have:
$$2l + 2w + 2h = 208 \qquad \text{(Equation 3)}$$
Now, we need to find the volume of the box when each dimension is increased by one inch, i.e., $l+1$, $w+1$, and $h+1$. Let's represent this new volume as $V_{\text{new}}$.
The volume of the new box can be calculated using the equation:
$$(l+1)(w+1)(h+1) = V_{\text{new}} \qquad \text{(Equation 4)}$$
To find $V_{\text{new}}$, we need to solve the system of equations formed by Equation 1, Equation 2, and Equation 3. Once we find the values of $l$, $w$, and $h$, we can substitute them into Equation 4 to find $V_{\text{new}}$.
Let's proceed to solve the equations to find $l$, $w$, and $h$.