As explained in Lickorish`s book "Introduction to knot theory", one can define the Conway-normalized version of the Alexander polynomial by the determinant of certain sum of Seifert matrix plus transposed Seifert matrix.
So let be $K$ a tame knot in $S^3$, Let $F$ be a Seifert surface and $A$ a Seifert matrix for $K$, which is defined by the evaluation of the Seifert form on $H^1(F;\mathbb{Z})\times H^1(F;\mathbb{Z})\to \mathbb{Z}$. Then $ \Delta_K:=\operatorname{det}(t^{1/2} A-t^{-1/2} A^T). $
So how does one calculate for the unknot? First homology group is trivial group, the tuple of basis vectors that one would insert in the Seifert form is the empty tuple. And then the determinant of an $(0\times 0)$-matrix is $1$. Is this correct?
It sounds strange to me. I know how one can compute $\Delta_{\text{unknot}}=1$ using the method based on the infinite cyclic cover of the knot complement, but wanted to avoid using two definitions.
As you explained, performing a computation with the disk requires you to interpret a bunch of definitions and conventions for a rank-zero module. Probably not a very enlightening exercise. But Lickorish's definition holds for any Seifert surface we choose, so we can avoid using the disk altogether:
Let $F$ be a torus with one boundary component, embedded in $S^3$ in the standard way. Then $H_1(F;\mathbb{Z})$ is generated by a longitude $a$ and meridian $b$. Depending on how you orient the curves, you'll get a Seifert matrix $A$ that looks like $$A=\begin{pmatrix} lk(a,a^+) & lk(a,b^+) \\ lk(b,a^+) & lk(b,b^+)\end{pmatrix} = \begin{pmatrix} 0 & \pm 1 \\ 0 & 0\end{pmatrix}.$$ Then we have $$\Delta_{\operatorname{unknot}}(t)=\det(t^{1/2} A - t^{-1/2} A^T)=\det\begin{pmatrix} 0 & \pm t^{1/2} \\ \mp t^{-1/2} & 0 \end{pmatrix}=0-(\pm t^{1/2})(\mp t^{-1/2})=1.$$