I don't know if the bond dimension is related to the spatial dimension of many-body systems.
As I understand it from this note, the bond dimension can be viewed as a 'refinement parameter', ie the set of states that can be represented by a matrix product state.
A matrix product state of 'bond dimension' $D$ (with periodic boundary conditions) is a pure state vector of the form:
$$
c_{j_1, \ldots, j_n}=\sum_{\alpha, \beta, \ldots, \omega=1}^D A_{\alpha, \beta ; j_1}^{(1)} A_{\beta, \gamma ; i_2}^{(2)} \ldots A_{\omega, \alpha ; j_n}^{(n)}=\operatorname{tr}\left(A_{j_1}^{(1)} A_{j_2}^{(2)} \ldots A_{j_n}^{(n)}\right)
$$
where the trace and the matrix product are taken over the contracted indices.
The remaining physical indices $j_1, \dots, j_n$ are open.
For example, if we take $D=1$, the matrices that we have in the above definition become complex numbers and we obtain a product state.
Small $D$ means that the set of states described by the MPS is 'refined'.
For larger bond dimensions, the set of states that can be represented becomes bigger and so the quality of the approximation of the natural states also.
Note however that a MPS already is a good approximation of natural states emerging in physical systems for not-so-big bond dimension $D$.
For open boundary conditions, the matrix $A^{(1)}$ is no longer a matrix from $\mathbb{C}^{D \times D}$, but $A^{(1)} \in \mathbb{C}^{1 \times D}$ is a row vector. Similarly, $A^{(n)} \in \mathbb{C}^{D \times 1}$ is a column vector. The expression of the MPS becomes
$$
c_{j_1, \ldots, j_n}=\sum^D A_{\alpha ; j_1}^{(1)} A_{\beta, \gamma ; i_2}^{(2)} \ldots A_{\omega ; j_n}^{(n)}=A_{j_1}^{(1)} A_{j_2}^{(2)} \ldots A_{j_n}^{(n)}
$$
but the idea remains the same.
