Given a vector space $V$ over field $\mathbb{C}$, define the Fock space $\mathcal{F}(V)$ as $$\mathcal{F}(V)\equiv \oplus_{N=0}^{\infty} V^{\otimes N} $$ with $V^{\otimes 0}\equiv\mathbb{C}$.
For example, if $V=\mathbb{C}$, then $V^{\otimes N}=\mathbb{C}$, so $\mathcal{F}(\mathbb{C})=ℓ^2(\mathbb{N})=L^2(\mathbb{R})$.
What's about $\mathcal{F}(\mathbb{C}^N)$? Is it $ℓ^2(\mathbb{N})\otimes ℓ^2(\mathbb{N}) \cdots \otimes ℓ^2(\mathbb{N}) $($N$ times tensor product), or $ℓ^2(\mathbb{N})\oplus ℓ^2(\mathbb{N}) \cdots \oplus ℓ^2(\mathbb{N}) $($N$ times direct sum)? Or $\mathcal{F}(\mathbb{C}^N)$ is neither of them.
First of all, keep in mind that all seperable infinite-dimensional Hilbert spaces are isomorphic. So any combination of tensors and sums that has a countable basis will be the right answer. I guess what you want to know is are they naturally isomorphic? Even $\ell^2(\mathbb{N})$ and $L^2(\mathbb{R})$ are not naturally isomorphic, right?
When you say that $\mathcal{F}(\mathbb{C})$ is $\ell^2(\mathbb{N})$, what isomorphism do you have in mind? Presumably one which maps $a_0 + a_1 e + a_2 e\otimes e + \dotsb$ (where $e$ is a basis for $\mathbb{C}$) to the $\ell^2$ sequence $(a_0,a_1,a_2,\dotsc)$. These spaces have canonical bases, so this is a natural isomorphism.
Similarly, for $\mathcal{F}(\mathbb{C}^n)$, the tensors are still finite dimensional, so the same thing works. We have a basis $e_1,\dotsc,e_n$ for $\mathbb{C}^n$, then a basis for eg $\mathbb{C}^n\otimes \mathbb{C}^n$ is $e_1\otimes e_1,e_1\otimes e_2,\dotsc$. Put these basis vectors in multi-index order, then we can map a vector $a_0 + a_1^ie_i + a_2^{ij}e_i\otimes e_j+\dotsb$ in $\mathcal{F}(\mathbb{C}^n)$ to a sequence of the coefficients in $\ell^2$. So here is an isomorphism $\mathcal{F}(\mathbb{C}^n)\cong\ell^2(\mathbb{N})$. But again, all separable infinite dimensional Hilbert spaces are isomorphic to $\ell^2(\mathbb{N}).$
In summary, a choice of basis for $V$ induces an enumeration $\mathcal{F}(V)$, and a choice of basis is more or less the same as an isomorphism to $\ell^2(\mathbb{N}).$ You may also wish to realize the isomorphism with $\ell^2(n^0)\oplus\ell^2(n)\oplus\ell^2(n^2)\oplus\dotsb$.
While it is also true that $\underbrace{\ell^2(\mathbb{N})\oplus\dotsb\oplus\ell^2(\mathbb{N})}_{n\text{ times}}$ and $\underbrace{\ell^2(\mathbb{N})\otimes\dotsb\otimes\ell^2(\mathbb{N})}_{n\text{ times}}$ are separable Hilbert spaces, and therefore isomorphisms could be proved to exist, I do not expect them to be natural. In particular, I do not think there is a natural isomorphism of functors like $\mathcal{F}(V\oplus W)\cong \mathcal{F}(V)\oplus\mathcal{F}(W)$ or $\mathcal{F}(V\oplus W)\cong \mathcal{F}(V)\otimes\mathcal{F}(W)$, which assumption seems to be the motivation of the question.