Let's consider the vector space $V_{\mathbb{K}, n} = \mathbb{M}_n(\mathbb{K})$ where $\mathbb{K}$ is a field.
We know that : $\dim(V_{\mathbb{R}, n}) = n^2$.
I would like to know if it's possible to have an explicit formula for : $\dim(V_{\mathbb{K}, n})$ ?
Fro example using the fact that for any $z \in \mathbb{C}$ whe have : $z = a+ib$ for some $a, b \in \mathbb{R}$, we have :
$$\dim(V_{\mathbb{C}, n}) = 2 \cdot n^2$$
So can we get an information $i$, on the field such that it's possible to have an explicit formula ?
The notion of "dimension" is field-specific. Thus $V_{\mathbb C,n}$ has dimension $2n^2$ over $\mathbb R$, but dimension $n^2$ over $\mathbb C$. What this means is that we can find a basis of $n^2$ vectors $\{v_1,v_2,\dotsc,v_{n^2}\}$ in $V_{\mathbb C,n}$ which are linearly independent over $\mathbb C$ (i.e., if $\sum_{i=1}^{n^2} \alpha_i v_i = 0$, where $\alpha_i$ are scalars in $\mathbb C$, then all $\alpha_i=0$) and span $V_{\mathbb C,n}$ (i.e., every vector $v \in V_{\mathbb C,n}$ can be expresseed as $v=\sum_{i=1}^{n^2} \alpha_i v_i$ for some scalars $\alpha_i \in \mathbb C$). If we restrict the scalars $\alpha_i$ in both definitions to belong to $\mathbb R$ only, we'll need $2n^2$ basis vectors. Thus we would say $\dim_\mathbb C V_{\mathbb C,n} = n^2$ but $\dim_\mathbb R V_{\mathbb C,n} = 2n^2$.
If $\mathbb K$ is a field extension of $\mathbb R$ (i.e., a field that contains $\mathbb R$), it can also be considered as a vector space over $\mathbb R$ and thus has a dimension over $\mathbb R$. If that dimension is finite, we say $\mathbb K$ is a finite field extension. In this case similar logic to what you presented in your question gives us that $\dim_\mathbb R V_{\mathbb K,n} = \dim_\mathbb R \mathbb K \cdot \dim_\mathbb K V_{\mathbb K,n} = \dim_\mathbb R \mathbb K \cdot n^2$. In the case of $\mathbb K = \mathbb C$, $\dim_\mathbb R \mathbb C = 2$ so we get $2n^2$ as you noticed.
If $\mathbb K$ is not a field extension of $\mathbb R$ -- e.g., if $\mathbb K = \mathbb Q$ -- then $V_{\mathbb K,n}$ will not be a well-defined vector space over $\mathbb R$.