7. As shown in Figure 1, in rectangle $A B C D$, $A B=3, A D=$ $4, E$ is a point on $A B$, and $A E=1$. Now, $\triangle B C E$ is folded along $C E$ so that the projection of point $B$ on plane $A E C D$ falls on $A D$. Then the volume of the pyramid $B-A E C D$ is
7. $\frac{4 \sqrt{3}}{3}$.

As shown in Figure 2, draw $B F \perp A D$, and connect $E F$,
By the problem, we know that $F$ is the projection of point $B$ on the plane $A E C D$. Therefore, $B F \perp$ plane $A E C D$.
Thus, $B F \perp E F, B F \perp C F$.
Let $B F=x, A F=y$. Then
$$
\begin{array}{l}
E F=\sqrt{y^{2}+1}, C F=\sqrt{(4-y)^{2}+3^{2}} \text {. } \\
\text { By } B F^{2}+E F^{2}=B E^{2}, B F^{2}+C F^{2}=B C^{2} \\
\Rightarrow\left\{\begin{array}{l}
x^{2}+\left(y^{2}+1\right)=2^{2}, \\
x^{2}+(4-y)^{2}+3^{2}=4^{2}
\end{array}\right. \\
\Rightarrow x=\frac{\sqrt{3}}{2}, y=\frac{3}{2} \text {. } \\
\end{array}
$$

Then the volume of the quadrilateral pyramid $B-A E C D$ is
$$
\begin{aligned}
V & =\frac{1}{3} S_{\text {quadrilateral } A E C D} \cdot B F \\
& =\frac{1}{3} \times \frac{1}{2}(1+3) \times 4 \times \frac{\sqrt{3}}{2}=\frac{4 \sqrt{3}}{3} .
\end{aligned}
$$