How does negative feedback change output impedance?
If I look into the output node (Vout) in the above figure (Figure 9.7 right), the unity gain feedback is pointing into the gate of M2 which is high impedance.

Not quite. Output impedance is a property of the circuit as a whole (how much current flows when a test voltage is applied, or conversely what voltage is seen when a test current is applied). When a voltage is applied, a current does arise from the output impedances of M2 and M4, but the transistors also amplify this applied voltage, causing a much larger current to flow at the output node (and hence a much lower output impedance).
Let's look at the output node and try to find the impedance seen there. We can do this by applying a small-signal voltage \$v_x\$ and noting the small-signal current drawn into that node as a result (call it \$i_x\$); at the same time we will keep the input at small-signal ground:

I'm going to simplify by taking \$r_{o2},r_{o4} \rightarrow \infty\$, on the basis that \$\frac{1}{g_m} \ll r_o\$.
The applied voltage causes a small-signal current of \$\frac{1}{2} g_m v_x\$ to flow in each branch:

M3 and M4 create a current mirror, which injects another small-signal current into the right branch:

and it is clear that for this simple 5-transistor unity gain buffer, the output impedance ends up being approximately \$\frac{1}{g_{m2}}\$.
Note further that as you take \$g_m \rightarrow \infty\$, gain tends to infinity, and output impedance tends to zero (i.e. with infinite gain, the amplifier's output voltage will not vary with a small-signal current injected into the output node)

(ro4 || ro2) which is the same as what it would be in an open-loop configuration.

Not quite, be careful to consider the behavior of the \$I_{ss}\$ tail current source. M2's source isn't connected to ground in the five-transistor OTA.