10 MHz frequency standard input

For a reduction in data rate due to interference that is as slight as that shown in your table you are very unlikely to see any deterioration of the spectrum whatsoever.

I am wondering what would be a good way to quantify this effect.

Do exactly what you did in the table - plot the effective data rate when an appliance is on compared to when it is off - the table tells you a lot so well done on that - it's useful to see. You could presume that the effective data rate reduction is due to packet retransmission due to packet corruptions. You could also assume that when data rates are higher, the packet length is shortened in order to make retries of smaller packets more liable to be received correctly - this also reduces overall data rate bearing in mind that the reduction would be greater if the packet length were kept bigger because of the probability of errors due to longer packets.

But why is the first entry in the FFT results not the fundamental?

Typically the 0'th entry stores the DC term. In some schemes, the [N/2]'th entry might be the DC term and the [N/2+1]'th entry would be the fundamental.

I applied an LC filter, by iterating through the FFT results and multiplying each bin by the filter gain at that frequency. I then did an inverse FFT, ... The result has both real and imaginary components.

Did you consider the phase effect of the filter? You should be multiplying by a complex value accounting for both the magnitude and phase response of the LC filter.

Even then, due to finite precision issues, you will often still get non-zero imaginary components in the inverse-transformed time-domain signal. But usually these are very small, for example on the order of \$10^{-15}\times\$ the real components. These are usually dealt with by simply ignoring them (taking the real part of the result only). Some tools even offer a built-in function (Mathematica's Chop, for example) to eliminate these artifacts.