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For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ is perfect and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

(I'll see if I can find an elegant way to define the abelianization map beyond the determinant part.)

For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.

For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ is perfect and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.

For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

(I'll see if I can find an elegant way to define the abelianization map beyond the determinant part.)

For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.

For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ is perfect and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

(I'll see if I can find an elegant way to define the abelianization map beyond the determinant part.)

For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.

For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

(I'll see if I can find an elegant way to define the abelianization map beyond the determinant part.)

For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ and hence the abelianization has order 2, given by the $\pm 1$ determinant.

For $n=2$, $-I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.

(I'll see if I can find an elegant way to define the abelianization map beyond the determinant part.)

For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.