Even for dimension $3$, a fundamental domain in the sense you describe will be quite hard to work with (Edit: writing it down actually isn't too bad, see the end of the answer). It's often simpler instead to to classify lattices not by a choice of basis but rather by their Gram matrix, namely the (symmetric, positive-definite) matrix of pairwise dot products. This information uniquely characterizes a basis up to orthogonal transformation, so you can recover a lattice with basis (up to rotation/reflection) from a Gram matrix.
To illustrate the usefulness of this approach, suppose we take $(x,y)$ in the set $F$ you defined above. The lattice with basis $(1,0),(x,y)$ will have a Gram matrix $\begin{pmatrix} 1 & x \\ x & x^2+y^2\end{pmatrix}$. From this we can see that for lattices coming from $F$ (and their scalar multiples), their Gram matrices $\begin{pmatrix} a & b \\ b & c\end{pmatrix}$ satisfy certain linear inequalities in the entries: $0\leq 2b\leq a\leq c$. Contrast that with the quadratic constraints on the basis vectors themselves.
One generalization of this notion to arbitrary dimension is Minkowski reduction (see e.g. Donaldson's Minkowski Reduction of Integral Matrices section 2.4). Every lattice has a basis whose Gram matrix is Minkowski-reduced, and while the basis itself will not be unique (automorphisms of the lattice preserve the Gram matrix), if there exists a basis for which all the inequalities are strict, then the Gram matrix is uniquely determined up to replacing some basis vectors $b_i$ with $-b_i$. So the space of Minkowski-reduced Gram matrices gives a complete classification of lattices in $\mathbb{R}^n$ modulo orthogonal transformation, up to negating basis vectors and some issues involving the boundaries of the region.
Unfortunately, even writing down the set of all necessary inequalities is hard in large dimensions. But it is doable in small dimensions. Let $A=\begin{pmatrix} a_{11}&\cdots& a_{1n}\\ \vdots & & \vdots \\ a_{n1} & \cdots & a_{nn} \end{pmatrix}$ be a Gram matrix.
Let $n\leq 4$. Then $A$ is the Gram matrix of a Minkowski-reduced basis if and only if the following conditions all hold: for each $k=1,\ldots n$ and each vector $\mathbf{x}\in \{-1,0,1\}^n$, if at least one of $x_k,\ldots,x_n$ is nonzero, then we have $\mathbf{x}^T A\mathbf{x}\geq a_{kk}$.
For instance, say $n=2$: from $k=1$ and $\mathbf{x}=(0,1)$ we get the condition $a_{22}\geq a_{11}$, and from $k=2$ and $\mathbf{x}=(1,\pm 1)$ we get the condition $a_{11}\pm 2a_{12}+a_{22}\geq a_{22}$, which implies $|2a_{12}|\leq a_{11}$. (All other $k$ and $\mathbf{x}$ end up giving redundant conditions.) By replacing $b_2$ with $-b_2$ if necessary, we can ensure $a_{12}\geq 0$, and so this exactly recovers the conditions described above.
If we do the same for $n=3$, we obtain the conditions $$0<a_{11}\leq a_{22} \leq a_{33},\qquad |2a_{23}|\leq a_{22},\qquad |2a_{12}|,\, |2a_{13}|\leq a_{11},$$ $$a_{12} +a_{13}-a_{23},\, a_{12} -a_{13}+a_{23},\, -a_{12} +a_{13}+a_{23},\, -a_{12} -a_{13}-a_{23} \leq\frac12(a_{11}+a_{22})$$ (see e.g. Rousseau's On Gauss's proof of Seeber's Theorem equations (1)-(3)). As before, we can negate $b_2$ and/or $b_3$ if necessary to ensure $a_{12},\,a_{23}\geq 0$; these inequalities then define a fundamental domain for the space of lattices.
Edit: Going from here to a region expressed in terms of basis vectors is straightforward. First we have to account for scaling and rotation: we can always transform the first basis vector to $(1,0,0)$, and then by rotation around the $(1,0,0)$ axis we can make the second basis vector take the form $(v,w,0)$. The third basis vector then has the form $(x,y,z)$, and we can compute the Gram matrix $$A=\begin{pmatrix} 1 & v & x \\ v & v^2 + w^2 & v x + w y \\ x & v x + w y & x^2 + y^2 + z^2 \end{pmatrix}.$$ Then we just need to ensure this basis is Minkowski-reduced. In other words, a fundamental domain for the space of lattices in $\mathbb{R}^3$ is given by $(v,w,x,y,z)\in\mathbb{R}^5$ satisfying the following conditions: $$1\leq v^2+w^2 \leq x^2+y^2+z^2,$$ $$0\leq 2(vx+wy)\leq v^2+w^2,\qquad 0\leq 2v\leq 1,\qquad -1\leq 2x\leq 1,$$ \begin{align*} v+x-(v x + w y),\, v -x+(v x + w y) &\leq\frac12(v^2+w^2+1),\\ -v +x+(v x + w y),\, -v-x-(v x + w y) &\leq\frac12(v^2+w^2+1). \end{align*} The fact that this fundamental domain is 5-dimensional means that unfortunately it doesn't have a nice "picture" like the space of 2-d lattices does. In general, the fundamental domain for lattices in $\mathbb{R}^n$ modulo scaling and orthogonal transformations will have dimension $\binom{n+1}{2}-1$. Even though this region does exist, in practice it's much easier to work with the Gram matrices because there are a lot more tools available to work with linear inequalities than with nonlinear inequalities.
