The real Clifford algebra $\text{Cl}_{0,4}(\mathbf{R})$ is isomorphic to $\text{M}_2(\mathbf{H})$, the algebra of $2\times 2$ matrices of quaternions.
In $\text{M}_2(\mathbf{H})$, how can I find a nice expression for the generators $e_1,e_2,e_3,e_4$ ?
Additionally, I appreciate any assistance on how to find expressions for generators of $\text{Cl}_{0,5}(\mathbf{R})\cong\mathbf{M}_4(\mathbf{C})$, $\text{Cl}_{0,6}(\mathbf{R})\cong\mathbf{M}_8(\mathbf{R})$, $\text{Cl}_{0,7}(\mathbf{R})\cong\mathbf{M}_8(\mathbf{R})\oplus\mathbf{M}_8(\mathbf{R})$ and $\text{Cl}_{0,8}(\mathbf{R})\cong\mathbf{M}_{16}(\mathbf{R})$.
José Figueroa-O'Farrill's notes (2.3 Filling in the Clifford chessboard) provide the answer. He establishes the recursion relation $\text{Cl}_{0,n+2}(\mathbf{R})\cong \text{Cl}_{n,0}(\mathbf{R})\otimes\text{Cl}_{0,2}(\mathbf{R})$.
Here $\mathbf{M}_2(\mathbf{R})\cong\text{Cl}_{0,2}(\mathbf{R})$ can be generated by
$e_1=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$, $e_2=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, and we have $e_1 e_2 =\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}$.
And for $\mathbf{H}\cong\text{Cl}_{2,0}(\mathbf{R})$, the quaternions, we can choose generators $e_3=i$ and $e_4=j$, with $e_3e_4=k$.
The recursion relation is established by the map $\phi(e_1)=\mathbf{1}\otimes e_1$, $\phi(e_2)=\mathbf{1}\otimes e_2$, $\phi(e_3)=e_3\otimes e_1 e_2$, $\phi(e_4)=e_4\otimes e_1 e_2$.
This yields for $\mathbf{M}_2(\mathbf{H})\cong\text{Cl}_{0,4}(\mathbf{R})$ the generators $e_1=\begin{pmatrix}1 & 0\\0 & -1\end{pmatrix}$, $e_2=\begin{pmatrix}0 & 1\\1 & 0\end{pmatrix}$, $e_3=\begin{pmatrix}0 & i\\-i & 0\end{pmatrix}$, $e_4=\begin{pmatrix}0 & j\\-j & 0\end{pmatrix}$
Generators for the other Clifford algebras can be calculated with the help of two additional recursion relations
$\text{Cl}_{n+2,0}(\mathbf{R})\cong \text{Cl}_{0,n}(\mathbf{R})\otimes\text{Cl}_{2,0}(\mathbf{R})$
$\text{Cl}_{s+1,t+1}(\mathbf{R})\cong \text{Cl}_{s,t}(\mathbf{R})\otimes\text{Cl}_{1,1}(\mathbf{R})$