= Solution
Choose an infinitesimal <gauge transformation> $\delta A_\mu^a=D_\mu^{ab}\omega^b$, where $D_\mu^{ab}=\delta^{ab}\partial_\mu+g f^{acb}A_\mu^c$. Varying the axial gauge function at fixed $\sigma$ gives
$$
\delta f^a(x)=n^\mu D_\mu^{ab}\omega^b(x),\qquad
\boxed{M^{ab}(x,y)=n^\mu D_\mu^{ab}(x)\delta^{(4)}(x-y).}
$$
This is the <Faddeev-Popov operator>. The <Faddeev-Popov determinant> has a Grassmann representation $\det M\propto\int\mathcal D\bar c\,\mathcal Dc\,e^{i\int\bar cMc}$. Choosing this normalization for the antighost gives
$$
\boxed{\mathcal L_{\rm gh}=\bar c^a n\cdot\partial c^a
+g f^{acb}\bar c^a n^\mu A_\mu^c c^b.}
$$
A simultaneous antighost rephasing changes the overall ghost-action sign and vertex conventions, but not the determinant's field-dependent content. The essential facts are the kinetic operator $n\cdot\partial$ and a ghost-gluon vertex proportional to $n^\mu$.
Now average the fixed-$\sigma$ constraint with the stated Gaussian:
$$
\int\mathcal D\sigma\,\delta[n\cdot A-\sigma]
\exp\left[-\frac{i}{2\xi}\int d^4x\,\sigma^a\sigma^a\right]
=\exp\left[-\frac{i}{2\xi}\int d^4x\,(n\cdot A^a)^2\right]
$$
up to a field-independent normalization. Consequently
$$
\boxed{\mathcal L_{\rm gf}=-\frac1{2\xi}(n^\mu A_\mu^a)^2.}
$$
The <Faddeev-Popov ghost> determinant is not generally field independent before imposing the strict axial condition: its interaction term contains $n\cdot A$. In the $\xi\to0$ gauge this quantity vanishes on the integration slice, giving the decoupling established next. Defining the inverse of $n\cdot\partial$ still requires boundary conditions for residual gauge transformations.
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