Constructing a list of functions/formulas which describes a set of grid points in a 3D matrix
Yes.  You can.  You can use linear algebra and linear programming to find a $F$ that is true for all points in $S$ and false for all points in $S'$.

Small simplifications.
Note that $d$ is unnecessary.  $N$ is a constant, so without loss of generality we can assume $d=0$ (otherwise replace $e$ by $dN+e$).  So, we will assume each $f$ has the form $ai+bj+ck+e \bowtie 0$.  Also note that the contents of the matrix itself are irrelevant and all that matters is the set $S$.  Finally, I will assume you are looking for integer values of $a,b,c,e$, with no limit on how large or small they are.

Finding a $f$.
Let's start by showing how, given $S$, you can find a $f$ that is true for all points in $S$.  There are several cases, according to which relational operator $f$ uses:


Equality: For $f$ of the form $ai+bj+ck+e=0$, consider the $a,b,c,e$ as four unknowns.  For each point $(i,j,k) \in S$, we obtain a linear equation on these four unknowns.  You can use linear algebra to find a value for $a,b,c,e$ such that $ai+bj+ck+e=0$ is true for all $(i,j,k) \in S$, if one exists: basically, you use Gaussian elimination to invert the matrix resulting from this system of equations.  Moreover, you can find a basis for the set of all $a,b,c,e$ that are valid solutions.

In this way, we either find zero, one, two, three, or four independent equality constraints $f$ that will be true for all points in $S$.
Less than or equal to: For $f$ of the form $ai+bj+ck+e \le 0$, we can use linear programming to find $a,b,c,e$ that make this inequality true for all $(i,j,k) \in S$.  Each $(i,j,k) \in S$ yields a single linear inequality on the unknowns $a,b,c,e$.  Form a system of inequalities obtained by all the points in $S$.  Solving this then gives us a valid combination of values for $a,b,c,e$.  There may be many solutions; linear programming will give you an arbitrary solution, not necessarily the best one.

You can improve this to check whether there exists a single linear inequality that separates $S$ from $S'$: simply add the linear inequality $ai+bj+ck+e \ge 1$ for each $(i,j,k) \in S'$ to the system of linear equations above, and check whether it has a valid solution.  If yes, then you have a single linear inequality $f$ that yields a minimal solution (an $F$ containing only a single $f$), and you are done.  If not, next try the following heuristic to find a $f$ that separates $S$ from $S'$ as well as possible:

Introduce unknowns $x_{i,j,k}$ for each $(i,j,k) \in S'$, and constrain them by $0 \le x_{i,j,k} \le 1$.  Add constraints $ai+bj+ck+e \ge x_{i,j,k}$ for each $(i,j,k) \in S'$.  Now use linear programming to search for values for $a,b,c,e$ that maximize the objective function $\sum_{(i,j,k) \in S'} x_{i,j,k}$.  This will force $f$ to be true for all points in $S$ and aim to make it false for as many points in $S'$ as possible.

If you want $a,b,c,e$ to be integers, use integer linear programming instead of linear programming.
Less than: For $f$ of the form $ai+bj+ck+e &lt; 0$, we can adjust the above approach by using linear inequalities of the form $ai+bj+ck+e &lt; 0$ (or $ai+bj+ck+e \le -1$ if we require $a,b,c,e$ to be integers) and carrying on as before.
Inequality: Simply swap the roles of $S$ and $S'$ and then use the approach for equalities.


Finding $F$.
So, given sets $S,S'$, this shows how to find a single constraint $f$ that could be included in $F$.  But how do we find a full $F$?

I don't know of any efficient way to find the optimal $F$.  However, I can suggest a heuristic that I suspect will work well: use an iterative greedy algorithm, where at each iteration you find a $f$ that is true for all points in $S$ and that is false for as many points that haven't been covered yet as possible.

In particular, at each stage, we will have a candidate $F$ that is true for all points in $S$, and is false for some points in $S'$.  Define $T = \{(i,j,k) \notin S : F(i,j,k)=\text{True}\}$.  Our goal will be to find $f$ that is true for all points in $S$ and is false for as many points in $T$ as possible; then we will replace $F$ with $F \land f$.

Given $S,T$, how can we find $f$ that is true for all $S$ and false for as many points in $T$ as possible?  Well, I've already described how.  For each relational operator $\bowtie \in \{=,\ne,&lt;,&gt;\}$, use the procedures above to find a candidate $f$ that uses that operator and is true for all of $S$ and false for as much of $T$ as possible.  This gives you four candidates for $f$.  For each candidate, count how many elements of $T$ it is false for, and keep the best one you find.

In this way, I expect you'll fairly rapidly converge to a $F$ that is either optimal (contains the minimal number of clauses) or is close to optimal.