Is the dynamic resistance ignored in this output impedance formula?
Yes you are right the dynamic resistance of \$V_{be}\$ is neglected because it is usually very small in comparison with rest of the resistances in the circuit. If you write the exact equation (neglecting the output resistance of BJT), you will get:
$$V_{in} - \frac{I}{\beta}r_{source} - \frac{\beta + 1}{\beta}I(\frac{\alpha}{g_m}) = V_{out}$$
where \$\frac{\alpha}{g_m}\$ is the dynamic resistance of \$V_{be}\$
Thus,
$$V_{out} - V_{in} = -\frac{I(\beta + 1)}{\beta}(\frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m})$$
Since, \$Z_{out} = (V_{out} - V_{in})/(\frac{-I(\beta + 1)}{\beta})\$. Finally,

$$Z_{out} = \frac{r_{source}}{\beta + 1} + \frac{\alpha}{g_m}$$
To get the sense of numbers, usually \$\beta\$ ~ 100 so, \$\alpha = \frac{\beta}{\beta + 1}\approx 1\$. 
\$\frac{1}{g_m} = \frac{V_T}{I_c}\$, at room temperature \$V_T \approx 25mV\$ and assuming \$I_c\$ ~ \$1mA\$, \$\frac{1}{g_m}\$ ~ \$25\Omega\$.
Thus last term in the expression for output impedance is of the order of few ohms.
 The source resistance, on the other hand, is the coming from the output impedance of a previous amplifier stage and is usually high ~ \$10^5\Omega -10^6\Omega\$. So usually the first term dominates. So we can approximate the output impedance as:
\$Z_{out} \approx \frac{r_{source}}{\beta + 1} \approx \frac{r_{source}}{\beta}\$.