Solution:
(1) From the problem, we have $\begin{cases} 4a = 4\sqrt{2} \\ b = c \end{cases}$,
and since $a^2 = b^2 + c^2$, we solve to get $a = \sqrt{2}$, $b = 1$, $c = 1$,
thus, the equation of ellipse $C$ is $\boxed{\frac{x^2}{2} + y^2 = 1}$.

(2) When the slope of line $l$ does not exist, according to the problem, the equation of $l$ is $x=0$,
at this time, $E$, $F$ are the top and bottom vertices of the ellipse, and $|EF| = 2$,
since point $D$ always lies inside the circle with diameter $EF$ and $m > 0$,
we have $0 0$,
let $E(x_1, y_1)$, $F(x_2, y_2)$, then $x_1 + x_2 = \frac{-4km}{2k^2 + 1}$, $x_1x_2 = \frac{2m^2 - 2}{2k^2 + 1}$,
let the midpoint of $EF$ be $G(x_0, y_0)$,
then $x_0 = \frac{x_1 + x_2}{2} = \frac{-2km}{2k^2 + 1}$, $y_0 = kx_0 + m = \frac{m}{2k^2 + 1}$,
thus $G\left( \frac{-2km}{2k^2 + 1}, \frac{m}{2k^2 + 1} \right)$,
thus $|DG| = \sqrt{\left( \frac{-2km}{2k^2 + 1} \right)^2 + \left( \frac{m^2}{2k^2 + 1} + m^2 \right)} = \frac{m \sqrt{4k^2 + 12k^2 + 4}}{2k^2 + 1}$,
$|EF| = \sqrt{1 + k^2} \cdot \sqrt{(x_1 + x_2)^2 - 4x_1x_2} = 2\sqrt{2} \cdot \sqrt{1 + k^2} \cdot \frac{\sqrt{2k^2 + 1 - m^2}}{2k^2 + 1}$,
since point $D$ always lies inside the circle with diameter $EF$,
$|DG| 0$, we solve to get $0 < m < \frac{\sqrt{3}}{3}$,
thus, the range of $m$ is $\boxed{(0, \frac{\sqrt{3}}{3})}$.