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Basis $M=\{(3,1),(2,1)\}$. I solved that the dual basis \begin{equation} M^*=\{e^1,e^2\}=\{(1,-2),(-1,3)\} \end{equation}

Then I solved that matrix

\begin{equation} g=\left(\begin{array}{cc}1&-5\\-1&10\end{array}\right) \end{equation}

Now, my notes says that antisymmetrization matrix for purely covariant tensor (which is a bilinear form) is given as

$$\frac{1}{q!}\sum_{\pi \in S_q}\text{sgn}\pi.(e_{\pi(1)},.....,e_{\pi(n)})$$

Here $q$ is 2 (because bilinear form is a (0,2) tensor and since we're in 2 dimensional space, there should be two permutations).

If I understant correctly, my matrix should be equal to

$$\frac{1}{2}\left((e_1,e_2)-(e_2,e_1)\right),$$

I just don't know what these $e_1$ and $e_2$ mean (I suppose they are not meant to be the same as the basis vectors of $M$).

Are $e_1$ and $e_2$ meant to be columns of the matrix $g$?

In that case the matrix of antisymmetrization would be

$$\pi_A(g)=\frac{1}{2}\left(\begin{array}{cc}6&-6\\-11&11\end{array}\right)$$

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Actually, the $e_{\pi(i)}$ are inteded to be the same you computed in the dual basis, but with another meaning. In fact, $(e_2,e_1)$ and $(e_1,e_2)$ are to be intended as a permutation operators on the tensor.

In particular, $(e_1,e_2)$ is the identity operator, whereas $(e_2,e_1)$ swaps the indices as $$ (e_2,e_1) (e_1\otimes e_1) =e_2 \otimes e_2,\\ (e_2,e_1) (e_1\otimes e_2) =e_2 \otimes e_1,\\\dots $$ Notice that these operators are linear, and you can compute their effect on the matrix. In fact, $(e_1,e_2)$ does not change the matrix $g$, whereas $(e_2,e_1)$ coincides with the transposition of the matrix.

In your case, $$ \pi_A(g)=\frac{1}{2}(g-g^T) $$ that is what is commonly known as the skew(or anti)-symmetric part of $g$.

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