Since the area $BCD=80=\frac{1}{2}\cdot10\cdot16$, the perpendicular from $D$ to $BC$ has length $16$.
The perpendicular from $D$ to $ABC$ is $16 \cdot \sin 30^\circ=8$. Therefore, the volume is $\frac{8\cdot120}{3}=\boxed{320}$.

1. Identify the given information:
- The angle between faces $ABC$ and $BCD$ is $30^\circ$.
- The area of face $ABC$ is $120$.
- The area of face $BCD$ is $80$.
- The length of edge $BC$ is $10$.

2. **Calculate the height of triangle $BCD$ relative to side $BC$:**
- The area of triangle $BCD$ is given by:
$$\text{Area}_{BCD} = \frac{1}{2} \times BC \times \text{height}_{BCD}$$
- Plugging in the known values:
$$80 = \frac{1}{2} \times 10 \times \text{height}_{BCD}$$
- Solving for $\text{height}_{BCD}$:
$$\text{height}_{BCD} = \frac{80 \times 2}{10} = 16$$

3. **Determine the altitude of the tetrahedron to face $ABC$:**
- The height of the tetrahedron from vertex $D$ perpendicular to face $ABC$ can be found using the sine of the angle between the faces $ABC$ and $BCD$.
- Given that $\sin 30^\circ = \frac{1}{2}$:
$$h_{ABC} = 16 \sin 30^\circ = 16 \times \frac{1}{2} = 8$$

4. Calculate the volume of the tetrahedron:
- The volume $V$ of a tetrahedron is given by:
$$V = \frac{1}{3} \times \text{base area} \times \text{height}$$
- Here, the base area is the area of face $ABC$, which is $120$, and the height is $h_{ABC}$, which is $8$:
$$V = \frac{1}{3} \times 120 \times 8 = \frac{1}{3} \times 960 = 320$$

The final answer is $\boxed{320}$