Will the DC offset of signal affect the slew rate in an op amp?
In general, when driving to max swing the slew rate is current-limited into a rated [pF] load.
( e.g. scope probe and coupling capacitance to ground.)

As long as the offset is in the linear range, it is not affected by DC offset.



Details

Slew rate is limited by two factors whichever is smaller;

The main reason for slew rate is current limit.


\$dV/dt=I_{max}/C_{load}\$ by either + or - current limiting, if asymmetric . 


However, we use rise time to estimate bandwidth, but if less than output driver current limit, then it is due to internal compensation capacitor and effective -3dB BW


\$ \dfrac{ΔV}{Δt}= \dfrac{ΔV}{t_r}, ~~~t_r = \dfrac{0.35}{f_{-3dB}}\$    


for Δt = 10% to 90% of ΔVmax.  



Proof of "0.35":   

An Op-Amp (OA) or oscilloscope can be modelled as a single-pole RC circuit.   

There is a relationship between asymptote \$T=RC \$ ,  rise time, \$t_r\$ and bandwidth, \$BW=f_{−3dB}\$.

I should point out that  for \$RC=T\$, only gives us the linear slope asymptote,  but not the slew rate 10~90%.      

\$T=~ 0 \text{ to } 64~\text{%} ~Vin = V_{in}(1-e^{-1})\$ for t=RC,

The RC step response is \$\dfrac{V_{out}}{ V_{in}} = (1 – e^{-\frac{t}{RC}})\$  .......... (1)     

From the above equation, we find the V ratios ( =10%,90%)    

10% = 0.105*T
 90% = 2.30  *T

Thus (10% to 90%) \$t_r = ~2.2 RC\$  ........ (1a)
from (2.30-0.105)T = 2.195T and T=RC      

For the RC model, \$RC=\dfrac{1}{2\pi \cdot BW }\$      .....(2)

Combining the equations (1a) &amp; (2),  we have:

10 to 90% \$t_R= \dfrac{0.35}{BW} =  \dfrac{2.2 }{ 2π \cdot BW}\$