The volume of a rectangular pyramid is $\frac{1}{3} A h$, where $A$ is the area of the base. So if your hopper were not truncated, you'd find the volume $A_1$ by taking the area of the top surface of its content (width x length, right?), multiply by the depth, dividing by $3$, and be done.
But the hopper is truncated. So you measure the volume $A_2$ of the truncated part and subtract it. That's computed by taking the area of the bottom of the hopper, times the "missing" height to the tip, and dividing by $3$.
Now the volume of your material is $A_1 - A_2$.
You can go a bit further, though. Suppose the width at the top of the hopper is $w_2$, and the width at the truncated bottom is $w_1$, and the vertical distance from top to bottom is $h$. Then when the hopper is filled to depth $d$ (where $d = 0$ means there's nothing in it, and $d = h$ means it's completely full), the width of the filled portion (assuming the top of it has been flattened out so that the fill really IS a truncated pyramid!) is
$$
w(d) = w_1\frac{h-d}{h} + w_2 \frac{d}{h}
$$
which you can simplify algebraically if you like. Similarly, if the length of the hopper at top and bottom are $k_2$ and $k_1$ (I don't like using "l" because it's easy to confuse with "1"), then you get
$$
k(d) = k_1\frac{h-d}{h} + k_2 \frac{d}{h}.
$$
The only question that remains for getting a full picture is "how tall is the cut-off tip of the hopper?" We can figure that out by setting $w(d)$ to $0$ and solving for $d$:
$$
0 = w_2\frac{d}{h} + w_1 \frac{h-d}{h}\\
0 = w_2\cdot d + w_1 \cdot (h-d)\\
0 = w_2\cdot d + w_1 \cdot h- w_1 \cdot d\\
(w_1 - w_2) \cdot d = w_1 \cdot h
d = \frac{w_1 \cdot h}{w_1 - w_2}.
$$
That "d" value will be negative; its absolute value (which is what we care about) will be
$$
h_c = \frac{w_1 \cdot h}{w_2 - w_1}.
$$
So the volume $A_2$ of the missing "tip" is
$$
A_2 = \frac{1}{3} (w_1 k_1) \frac{w_1 \cdot h}{w_2 - w_1}
$$
and the volume of the whole pyramid (if the tip were not cut off), for fill-height $d$, is
$$
\begin{align}
A_1
&= \frac{1}{3} w(d) k(d) (d+h_c) \\
&= \frac{1}{3} \bigl( w_1\frac{h-d}{h} + w_2 \frac{d}{h} \bigr)
\bigl( k_1\frac{h-d}{h} + k_2 \frac{d}{h} \bigr)
(d + \frac{w_1 \cdot h}{w_2 - w_1})\\
&= \frac{1}{3h^2 (w_2 - w_1)} \bigl( w_1 h + (w_2 - w_1) d\bigr)
\bigl( k_1 h+ (k_2-k_1)d \bigr)
(w_2 d + w_1 (h-d))
\end{align}
$$
and the total volume of a truncated pyramid of height $d$ is the difference between $A_2$ and $A_1$, i.e, $A_1 - A_2$. That is, as you'd expect, a cubic equation in $d$, because as $d$ grows, the width, height, and depth all grow linearly, and the volume is (roughly) a product of these.
I'm sure it simplifies a lot when you do out the algebra, but I'm leaving it in this form, because it's clear what the various terms all mean.
