Minimal polynomial with $f(\alpha)=0$

We see that the four roots of $f$ are $\pm\sqrt3\pm\sqrt2i$ (where the $\pm$ are independent). You can show this for instance by inserting and calculating, or by noting that your method to find $f$ gives the same polynomial for all four numbers.

Any field extension of $\Bbb Q$ with $\alpha$ in it will also contain $\alpha^3=-3\sqrt3+7\sqrt2i$, and thus both $\sqrt3$ and $\sqrt2i$ (technically you can stop here, because this already shows that $[\Bbb Q(\alpha):\Bbb Q]\geq 4]$). Therefore such a field also contains the three other roots of $f$. Which is to say, any polynomial over $\Bbb Q$ that has $\alpha$ as a root also has all the other three roots of $f$ as roots. So your $f$ is minimal.

It suffices to prove that $1,\alpha,\alpha^2,\alpha^3$ are linearly independent over $\mathbb Q$. Writing them in the basis $1,\sqrt{3},\sqrt{2}i,\sqrt{6}i$ of $\mathbb Q[\sqrt{3},\sqrt{2}i]$ we get: $$ \begin{pmatrix} 1 \\ \alpha \\ \alpha^2 \\ \alpha^3 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 2 \\ 0 & -3 & 7 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ \sqrt{3} \\ \sqrt{2}i \\ \sqrt{6}i \end{pmatrix} $$ The key point is that the matrix is invertible. You can prove this by row reduction or by computing its determinant and checking that it is not zero.