Covariant surface vector

On pg 74 of Dalarsson's Tensors, Relativity and Cosmology (The Integral theorems for tensor field chapter), the covariant surface vector was defined as: $$dS_k=\frac{1}{2}\epsilon_{kmn}dx^mdx^n=\frac{1}{2}\sqrt{g}e_{kmn}dx^mdx^n. \tag{10.41}$$

In the Descartes coordinates, where $$\sqrt{g}=1$$, the components of this vector are given by $$dS_1=dx^2dx^3, dS_2=dx^1dx^3 , dS_3=dx^1dx^2 . \tag{10.42}$$

However, I attempted calculating the components in (10.42), for instance $$dS_1=\frac{1}{2}(e_{123}dx^2dx^3+e_{132}dx^3dx^2)$$ which yielded $$dS_1=\frac{1}{2}(dx^2dx^3-dx^3dx^2)$$ which was clearly incorrect. So it would be great if anyone could explain the errors that I have made.

The symbols $$dx^1$$ and $$dx^2$$ are presumably meant to be differential forms, and these anticommute: $$dx^1\wedge dx^2 =- dx^2\wedge dx^1.$$ Many authors leave out the $$\wedge$$ symbol when writing products of forms, and this (again "presumably" because I have not seen the book) is what the author is doing here. With this understood, your computation is essentially correct. When you insert infinitesimal vectors $$\delta {\bf x}_1$$ and $$\delta {\bf x}_2$$ into the two-form $$dS_3$$ you have $$dS_3 (\delta {\bf x}_1, \delta {\bf x}_2)=\frac 12 (dx^1\wedge dx^2-dx^2\wedge dx^1)(\delta {\bf x}_1, \delta {\bf x}_2)= \frac 12 (\delta x_1^1 \delta x_2^2- \delta x_1^2 \delta x_2^1) = \frac 12 (\delta {\bf x}_1 \times \delta {\bf x}_2)_3$$ which is the third component of the vector products of the displacements $$\delta {\bf x}_1$$ and $$\delta {\bf x}_2$$. This is the third component of the vector element of area defined by the parallelogram whose sides are $$\delta {\bf x}_1$$ and $$\delta {\bf x}_2$$.
Equation 10.41 can't be describing the area covector in terms of infinitesimal coordinate changes. If it was, then due to the antisymmetry of the Levi-Civita tensor $$\epsilon_{kmn}$$ on the indices $$m$$ and $$n$$, this expression would vanish identically.
So I think their notation $$dx^m dx^n$$ actually means the wedge product $$dx^m\wedge dx^n$$, i.e., these are differential forms. So your final result looks fine, because $$dx^3\wedge dx^2=-dx^2\wedge dx^3$$, so it reduces to $$dS_1=dx^2\wedge dx^3$$.