You can notionally build as many stages as you want with a single amplifier, and AFAIR I have seen a 5 stage design implemented just to make the point BUT it becomes increasingly hard to "realise" (= construct) as you add stages around a single amplifier. To obtain the correct ratios of components requires increasingly precise component values and increasingly stable components. Capacitors are hard to get with extremely high precision and resistors are only slightly better. For a two stage or 3 stage design you can in most cases manage with 1% parts. Beyond that, the fun begins.
Note: "Pole" used generally here rather than saying "pole or zero as is applicable ..." in each case.
While you will notionally get the same result from a bandpass filter by cascading stages in any order, you will find that in limiting cases aspects such as stage Q and signal magnitude will have some effect. The same applies to stage order in a multiple stage low or high pass.
Your circuits are unusual in separately providing gain for the amplifier. This is acceptable, but the norm is to use a unity gain buffer in this application - amplifier Vout connected to amplifier inverting input. The addition of gain will also affect filter Q and you will end up not realising a classic filter polynomial if you alter the gain - assuming the designer implemented a 'proper' filter in the first place. In the case of the multipole design, varying the gain arbitrarily as shown will influence the "shape" of the resultant response rather than just its amplitude.
For one and two pole designs that need a unity gain buffer, you can use a 1 transistor emitter follower with usually acceptable results. As shown below, the results with a transistor with relatively low gain are inferior to results usually available from an opamp, but can still be very useful..
The above diagram is from this extremely good page -
Elliott Sound products: Active filters - Characteristics, Topologies, Examples
Lots more on the above, and related, here - Gargoyle search.
First, \$\alpha = 2 \zeta\$, \$\alpha\$ (damping constant) is used in filters, and typically \$\zeta\$ is used in controls.
Q is defined as \$ Q = \dfrac {f_o}{\Delta f}\ (= \dfrac{1}{\alpha}) \$. It is a band pass parameter and does not apply to first order filters.
If you want to change the step response, you're going to have to change the damping. In the case of the Sallen-Key, that involves changing the gain of the amplifier. This is where you have to start being careful. Damping is set based off the approximation you chose. As an example, for a Butterworth approximation \$\alpha = 1.414\$. If you want a faster rise time, you'll want a lower damping constant. Look at the Chebyshev =approximations. You do need to be aware that there will be overshoot with those approximations.
Cascading filters does not have the effect you think it does. For example, simply cascading two second order Butterworth low pass filters for example, does not equal a fourth order Butterworth low pass. There are tables of common approximations and the appropriate correction factors that need to be considered when cascading filters.
Best Answer
You can make a high-order low-pass filter with no op-amps at all. In principle you could obtain any roll-off rate you like, given enough pi or T sections. You could then add an op-amp output buffer to make your filter a "one op-amp" filter if for some reason you'd like to do that.
But high-order filters tend to be very sensitive to small changes in component values, so they often require hand-tuning each circuit to achieve good performance, and the performance could drift as temperature changes.
The benefit of using active filters is that the op-amp allows low output impedance and high input impedance, reducing the loading effect of one stage on the next. This means you can cascade several first or second order filters without much interaction from stage to stage, and reduce the need for tuning. If you try to make a high-order active filter in one stage, you're giving up a key advantage of using active filters.