Impedance for Helmholtz Coil Connected to Audio Amplifier

"since they are in parallel Xtot should decrease"

That's not automatically true for complex impedances. For the inductor \$Z_L = j\omega L\$, for the capacitor \$Z_C = \dfrac{1}{j \omega C}\$. Both in parallel gives

\$ Z = \dfrac{j\omega L \cdot \dfrac{1}{j \omega C}}{j\omega L + \dfrac{1}{j \omega C}} = \dfrac{j\omega L \cdot \dfrac{1}{j \omega C}}{\dfrac{j^2 \omega^2 LC + 1}{j \omega C}} = \dfrac{j\omega L}{1 - \omega^2 LC} \$

The minus sign in the denominator is interesting. Since \$\omega^2 LC\$ is positive, we can find a value for it where the denominator becomes zero.

\$ \omega^2 LC = 1\$

or

\$ \omega = \sqrt{\dfrac{1}{LC}} \$

If the denominator is zero the impedance is infinite. That goes against intuition which says that paralleling two components will give a lower impedance than the lowest of the two.

The image gives the explanation. Currents through \$C\$ and \$L\$ are at 90° with voltage, but in opposite directions. If their magnitudes are different their sum will be either a capacitive or inductive current. But when they're equal the sum is zero. No current. A zero current for a non-zero voltage means infinite impedance.

The frequency for which this is true is the resonance frequency, and it's used to make oscillators and high-Q filters.

In this case, you are being led astray by what you don't know. A speaker coil has a (more or less) constant impedance because it drives a cone against air resistance. In your case, with no speaker attached, it behaves as an inductor, not a resistor. From the numbers you provide, it will have an inductance of ~ 5 milliHenrys (.005 H). For an inductor, the current (and the magnetic field) will go down as frequency goes up. The relevant equation deals with impedance, not resistance, represented by Z.

                    Zl = 2 x pi x L x f

and this will equal 8 ohms at f = 250 Hz (about). So, for frequencies below about 250 Hz, the total impedance (sqrt(Zl^2 + R^2)) will increase with frequency, until at 250 Hz your current amplitude will be about 70% of its DC value. Further increases in frequency will continue to increase impedance and current will drop further as you increase the frequency. You probably don't need to hear about phase shifts.

The effect of all of this is to reduce power dissipation in your coil as frequency increases. But particularly at low frequencies you do need to keep a very close watch on your temperatures. Dissipating 800 watts in a coil the size you're considering stands a pretty good chance of melting the interior layers. It certainly nay cause the bobbin to melt or catch fire.