Let $\mathbb{H}$ denote the quaternions and $SL_n(\mathbb{H})$ (with $\mathfrak{sl}_n(\mathbb{H})$ its respective Lie algebra) the group of matrices with determinant 1 with entries over $\mathbb{H}$.
In order for the Lie bracket to be closed and therefore talk about the respective Lie algebra $\mathfrak{sl}(\mathbb{H})$ we will identify each element of $\mathbb{H}$ with their respective matricial representation and regard them as $2 \times 2$ matrices with complex entries. Therefore, each $SL_n(\mathbb{H})$ will be a $2n \times 2n$ matrix with complex entries and we can write
$$ SL_n(\mathbb{H}) = M_n(\mathbb{H}) \cap SL_{2n}(\mathbb{C})$$
I am trying to find the complexification $\mathbb{C} \otimes_{\mathbb{R}} \mathfrak{sl}_n(\mathbb{H})$. By construction we have an isomorphism $$ \mathbb{C} \otimes_{\mathbb{R}} \mathfrak{sl}_n(\mathbb{H}) \simeq \mathfrak{sl}_n(\mathbb{H}) \oplus i\mathfrak{sl}_n(\mathbb{H}) $$
The Lie algebra $\mathfrak{sl}_n(\mathbb{H}) $ will consist of traceless matrices in $M_n(\mathbb{H})$. I also know that $\mathbb{C} \otimes_{\mathbb{R}} \mathbb{H} \simeq M_2(\mathbb{C})$ because $\mathbb{H} \oplus i\mathbb{H} \simeq M_2(\mathbb{C})$.
I hunch that the complexification of the speciallinear quartenionic group will be $SL_{2n}(\mathbb{C})$ because of the remarks above, however I don't know if how to prove it nor do I even know it is true. Any help?
As complex associative algebras, $\mathbf{H}\otimes_\mathbf{R}\mathbf{C}\simeq M_2(\mathbf{C})$, whence $M_n(\mathbf{H})\otimes_\mathbf{R}\mathbf{C}\simeq M_n(M_2(\mathbf{C}))\simeq M_{2n}(\mathbf{C})$.
Hence, as complex Lie algebras $\mathfrak{gl}_n(\mathbf{H})\otimes_\mathbf{R}\mathbf{C}\simeq\mathfrak{gl}_{2n}(\mathbf{C})$. Passing to the derived subalgebra (which commutes to complexification) yields $\mathfrak{sl}_n(\mathbf{H})\otimes_\mathbf{R}\mathbf{C}\simeq\mathfrak{sl}_{2n}(\mathbf{C})$.