If a hypersurface in $\mathbb{R}^n$ is implicitly defined by the equation $F(x_1, \dots, x_n) = d$ for some $d \in \mathbb{R}$, then you can consider it as a level curve of the function $F$. As you mention, $\nabla F$ is normal to the surface, so, provided you can determine $F$, you can find the normal vector at a point on the surface.
In your situation, the surface is defined by an equation $F(x, y, z) = d$. Furthermore, it is a plane, so $F(x, y, z) = ax + by + cz$, where $a, b, c \in \mathbb{R}$ are yet to be determined. From the figure, we know that $(1, 0, 0)$, $(0, 2, 0)$ and $(0, 0, 3)$ are on the surface, and therefore must satisfy $F(x, y, z) = d$. Substituting in each point, one by one, we get:
\begin{align*} F(1, 0, 0) &= d \Rightarrow a(1) + b(0) + c(0) = d \Rightarrow a = d,\\ F(0, 2, 0) &= d \Rightarrow a(0) + b(2) + c(0) = d \Rightarrow 2b = d \Rightarrow b = \dfrac{1}{2}d,\\ F(0, 0, 3) &= d \Rightarrow a(0) + b(0) + c(3) = d \Rightarrow 3c = d \Rightarrow c = \dfrac{1}{3}d. \end{align*}
Now replacing $a, b,$ and $c$ in the equation for $F$, we obtain:
$dx + \frac{1}{2}dy + \frac{1}{3}dz = d.$
If $d = 0$, then every element $(x, y, z) \in \mathbb{R}^3$ would satisfy the equation and hence be on the surface. As this is not the case, $d \neq 0$ so we can divide both sides of the above equation by $d$, leaving us with:
$x + \frac{1}{2}y + \frac{1}{3}z = 1.$
If you like, you can multiply both sides by $6$ so that all the coefficients are integers, in which case the equation is $6x + 3y + 2z = 6$. Therefore $F(x, y, z) = 6x + 3y + 2z$, so $\nabla F = (\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}) = (6, 3, 2)$ is a vector normal to the surface. As $\|\nabla F\| = \sqrt{6^2 + 3^2 + 2^2} = \sqrt{49} = 7$, $\frac{1}{\|\nabla F\|}\nabla F = \frac{1}{7}(6, 3, 2)$ is a unit normal to the surface.
Note, as this surface is a plane in $\mathbb{R}^3$, you can obtain the normal in a quicker way.
If we can find two linearly independent vectors in the plane, then their cross product will be normal to the plane. We can use the points given to find two such vectors. As $(1, 0, 0)$ and $(0, 2, 0)$ are in the plane, the vector $(0, 2, 0) - (1, 0, 0) = (-1, 2, 0)$ from $(1, 0, 0)$ to $(0, 2, 0)$ is in the plane. Likewise $(0, 3, 0) - (1, 0, 0) = (-1, 0, 3)$ is also in the plane. Clearly these two vectors are linearly independent. Therefore:
$(-1, 2, 0) \times (-1, 0, 3) = \left| \begin{array}{ccc} i & j & k \\ -1 & 2 & 0 \\ -1 & 0 & 3 \end{array} \right| = 6i + 3j + 2k = (6, 3, 2),$
as before.
As for your concerns in higher dimensions, the approach I outlined is the standard approach, except that you usually know $F(x_1, \dots, x_n)$ and $d$, so all that you need to do is calculate $\nabla F = (\frac{\partial F}{\partial x_1}, \dots, \frac{\partial F}{\partial x_n})$. Most of the time $\nabla F$ will not be a constant vector as its components can be functions - in fact, the only time $\nabla F$ is constant is when your surface is a hyperplane in $\mathbb{R}^n$. Either way, $\nabla F$ is a vector which (can) vary from point to point; this is what we call a vector field. More formally, a vector field on a hypersurface $\Sigma$ is a function $\Sigma \to \mathbb{R}^n$; $\nabla F$ is such a function.