Constant voltage control via op amp and transistor
I think your original confusion is more significant than the specifics of LM317 or what not.

A transistor cannot manufacture current. The relationship \$I_c = \beta I_b\$ only holds when whatever that supplies \$I_c\$ can provide the current required.

Using your example, if you put a 1K ohm resistor at R3 then by the time the current gets to around 27mA, the collector of the transistor would drop to 1V and there is no way for the transistor to source more current to supply the 1V output. So when the load demands more than 27mA, the opamp+transistor can no longer keep up and output voltage drops below 1V (ignoring the current that can source through the base by the opamp).

The second misconception you seemed to have described is that \$V_c \approx V_b\$. From your description I think you got 1.1A from (28V - 1V - 0.6V)/25 regardless of load. The assumption of \$V_c \approx V_b\$ is simply not valid. \$V_c\$ will take on whatever voltage necessary while providing \$I_c\$ (subject to operating limits). Therefore, you can actually short out R3 and the circuit would work correctly unless you overload something.

As an example, with a load of 2 ohm and \$V_{out} = 1V\$, that determines \$I_e = 0.5A\$. \$I_c = I_e + I_b = \frac {\beta}{1+\beta} I_e\$ which is still approximately 0.5A. \$V_c\$ can very well sit at 28V directly with 0.5A of \$I_c\$. The power dissipation will be 0.5A x 27V = 13.5W.

Using the same example with the 25 ohm resistor at R3. Then there is a (\$I_c \times 25\$) voltage drop across R3 of 12.5V. Now \$V_c\$ is at 15.5V, still well within the operating mode of transistor. The 13.5W power dissipation is now distributed between the transistor and the 25 ohm resistor. With a 25 ohm resistor, you are indeed limited to no more than 1.1A of maximum output current at 1V output.

By the way, this is a very wasteful power supply. The maximum efficiency is 5V max / 28V. That is less than 20% efficiency at best.