1. There is a general algebraic result which states that the abstractly defined associative $\mathbb{R}$-algebra with generators $X_1,\ldots,X_n$, $Y_1,\ldots,Y_n$ and relations $$ X_iX_j+X_jX_i=0, \quad Y_iY_j+Y_jY_i=0,\quad X_iY_j+Y_jX_i=\delta_{i,j}1 $$ for all $i,j=1,\ldots,n$, is of dimension $2^{2n}$ and is isomorphic to the algebra of $2^n\times 2^n$-matrices. In fact, one can do better: those matrices are matrices of all possible linear transformations of the Grassmann algebra $\Lambda(Y_1,\ldots,Y_n)$, on which $Y_i$ acts by multiplication, and $X_i$ acts as the odd derivation $\frac{\partial\phantom{Y_i}}{\partial Y_i}$.

2. Since we are working with the matrix algebra, every module over it is a direct sum of several copies of the standard module. In particular, every module of dimension $2^n$ is isomorphic to the one given by the action on the Grassmann algebra. In particular, there is, up to rescaling, just one vector annihilated by all $X_i$.

3. All the questions in your post are addressed by this. Roughly speaking, you define a representation $\rho$ of this algebra by putting $\rho(X_i)=A_i$, $\rho(Y_i)=A_i^*$. The only thing to take care of is that you work over $\mathbb{C}$, not over $\mathbb{R}$, but this you can take care of by separating real and imaginary parts of your operators.

1. There is a general algebraic result which states that the abstractly defined associative $\mathbb{R}$-algebra with generators $X_1,\ldots,X_n$, $Y_1,\ldots,Y_n$ and relations $$ X_iX_j+X_jX_i=0, \quad Y_iY_j+Y_jY_i=0,\quad X_iY_j+Y_jX_i=\delta_{i,j}1 $$ for all $i,j=1,\ldots,n$, is of dimension $2^{2n}$ and is isomorphic to the algebra of $2^n\times 2^n$-matrices. In fact, one can do better: those matrices are matrices of all possible linear transformations of the Grassmann algebra $\Lambda(Y_1,\ldots,Y_n)$, on which $Y_i$ acts by multiplication, and $X_i$ acts as the odd derivation $\frac{\partial\phantom{Y_i}}{\partial Y_i}$.

To prove the latter point, one can argue as follows:

A) the operators of multiplication by $Y_i$ and odd derivatives $\frac{\partial\phantom{Y_i}}{\partial Y_j}$ satisfy these commuting relations, this is an easy calculation, so there is a surjective map from the abstract algebra A with the commutation relations $$ X_iX_j+X_jX_i=0, \quad Y_iY_j+Y_jY_i=0,\quad X_iY_j+Y_jX_i=\delta_{i,j}1 $$ to the algebra of "differential operators" on the Grassmann algebra (algebra generated by multiplications by $Y_i$ and odd derivatives $\frac{\partial\phantom{Y_i}}{\partial Y_j}$).

B) every endomorphism of the matrix algebra can be represented by a differential operator, since it is easy to express each matrix unit sending $Y_I=Y_{i_1}\wedge\cdots\wedge Y_{i_k}$ to $Y_J=Y_{j_1}\wedge\cdots\wedge Y_{j_p}$ and others to zero: first multiply by all the $Y$'s not in $I$, then take derivative with respect to each $Y_i$ once, then multiply by all the $Y$'s from $J$.

C) Thus, we have a surjective map from our algebra to the matrix algebra, but our commutation relation allow to put all $X$'s before $Y$'s and order them, which shows that the dimension of our algebra is at most $2^{2n}$, and therefore the surjective map we constructed is an isomorphism.

2. Since we are working with the matrix algebra, every module over it is a direct sum of several copies of the standard module. In particular, every module of dimension $2^n$ is isomorphic to the one given by the action on the Grassmann algebra. In particular, there is, up to rescaling, just one vector annihilated by all $X_i$.

3. All the questions in your post are addressed by this. Roughly speaking, you define a representation $\rho$ of this algebra by putting $\rho(X_i)=A_i$, $\rho(Y_i)=A_i^*$. The only thing to take care of is that you work over $\mathbb{C}$, not over $\mathbb{R}$, but this you can take care of by separating real and imaginary parts of your operators.