How do I analyze this negative feedback?
The problem you are facing stems from the fact that the ideal feedback relation
$$A_f=\frac{A_{ol}}{1+A_{ol} B} $$
is derived from block diagrams, and block diagrams have the peculiarity of unilaterally transferring signals. Block diagrams do not model load effects or the inherent bidirectionality of power transfer enacted by two-ports. And two-ports - which relate pairs of variables (voltage, current) at the input and output are what we naturally use to solve circuits.
When we solve a feedback circuit using two-ports we implicitly take into account bidirectionality and loading effects and that might introduce spurious terms in the feedback relation.
Since this feedback configuration samples the output voltage and compares the input current, it is best explained by using an input current generator (as you ask in the comments above). So, here is your circuit with an input current source

Once we have solved biasing and found the values of the small signal parameters, we can solve it with a reasonably detailed model for the BJT (I assume that the input capacitor is there for decoupling purposes, and since it is in series with an ideal current source I will neglect it in the analysis - simulation with an ideal current source confirms it won't affect the output)

KCL at node 1 says

KCL at node 2 says

by eliminating unneeded variables and after a bit of algebraic massaging we get

an expression that we can recast as

Note what happens if we remove the feedback by making \$R_f\$ go to infinity: \$R_c/R_f\$ goes to zero and \$r_\pi //R_f\$ becomes \$r_\pi\$, while in the denominator the whole middle term is turned into nothingness. Hence the open loop gain becomes

Now, let's get back to the complicated feedback relation we have found above and let's see what happens when we choose the feedback network in such a way that it is the least disturbing as possible (while still performing its function).
If \$R_f\$ is much bigger than \$R_c\$ we can neglect the \$R_c/R_f\$ term and approximate the parallel of \$r_\pi\$ and \$R_f\$ with just \$r_\pi\$. Being \$R_f\$ still finite, the middle term in the denominator won't go to zero, though. We get an approximate feedback relation that can be cast in the form we have derived with block diagrams:

where

Note that I could have spared a bit of algebraic mess, had I chosen to use a BJT model with current control (\$i_c = \beta i_b\$ would have avoided bringing \$v_\pi\$ and \$r_\pi\$ along), and chosen the opposite conventional sign for \$i_f\$ (in that case the block diagram would have has a + summing node and the ideal formula would have been
$$A_f=\frac{A_{ol}}{1-A_{ol} B} $$
and we would have obtained a positive B.
Moreover, had I realized from the start that I wanted to avoid loading effects, I could have used a simplified and idealized version of the two port representing the amplifier stage with a zero input resistance (what we ideally want in an amplifier that accepts an input current) and a zero output resistance (what we really want in amplifier that produces a voltage output - note the by not including \$r_o\$ we already had that simplification). The analysis would give directly the simplified formula.