@Confused: Sorry - but I have to start with some general comments:
In principle, for an 8th order filter you have two basic alternatives:
(a) Direct realization (active topology derived from a passive and tabulated reference structure, and (b) Cascade realization as a series connection of 4 second-order stages.
I suppose, you are following the latter approach - and here you have again several alternatives (how the various 2nd order stages are realized). It seems that you have decided to use Sallen-Key realizations because you have mentioned finite gain values.
But also in this case, you again have alternatives: Unity gain approach, gain-of-two approach or equal-component approach (with gain values lower than "3"). Independent on these 3 alternatives, you must know that all 4 stages look different: They are individually designed for equal pole frequencies (applies only for Butterworth response) but for different pole-Q values to be found in filter tables. Hence, you will NOT have 4 identical 2nd-order stages but each of the 4 must bedesigned separately.
I am not sure if this answers all of your questions - perhaps it helps if you could give us some more detail of your envisaged design.
EDIT 1: The following link leads you to a document (from TI) which gives you the Q values for your 8th-order filter on page 8 (correction: page 9)
http://www.ti.com/lit/an/sloa049b/sloa049b.pdf
EDIT 2: For your convenience, here are the formulas for designing the 4 stages (equal pole frequencies wp, different Qp values) - to be applied for the gain-of-two version:
C1=C2=C and wp=1/[C*Sqrt(R1R3)] and Qp=Sqrt(R1/R3) with R1: Most left resistor(connected to input signal). For a gain of "2" you can use any two equal resistors in the negative feedback path.
The basic problem is that your half rail generator is being dragged down to less than 2.5 volts by the dc offset on your source voltage. Then you have what appears to be a gain of 2 opamp circuit but the interactions of the source voltage superimposed on what is meant to be a steady 2.5 volts is screwing things up. Replace the two resistors of 100 ohm with a fixed dc voltage source is my advice.
By the way, with a gain of nearly 2 the opamp dc output voltage will be naturally centred a little lower than 4 volts and this, I believe is not what you want.
Best Answer
Your LPF with 10Hz cutoff has transient reponse that will take about 0.5s (about 5*1/fc) to fully settle. That's what you are seeing.
For faster step-response settling you need to pick a wider filter with higher Fc.
Have a look here for the relationship between rise time and filter Fc:
Image from https://www.thorlabs.com/newgrouppage9.cfm?objectgroup_id=9817
There are circuit tricks to get it to start at a desired known non-zero DC by initializing your filter to a desired "reset" state. Although this does not change the rise time, it does allow the circuit to start at a pre-fixed DC level immediately after a reset. (This requires additional analog switches and control circuitry)