Not sure if this will answer your queries, but I did get intrigued by the implied question:
Could Newton have derived his theory of gravitation from dimensional analysis?
Newton had a hunch that planetary motion is nothing more than a direct result of the same effect that makes objects drop to earth: the presumed universal attraction between all masses. He wanted to derive the consequences of this hypothesis. Available to him were his three laws of motion, as well as two important observations:
1) the empirical fact that different masses undergo the same acceleration (observed by Galileo)
2) Kepler's third law of planetary motion: the square of the orbital period of a planet is proportional to the cube of the radius of its orbit around the sun.
The first observation tells Newton he has to consider the acceleration $g = F_g/m$ ($m$ being the planetary mass) due to gravity rather than the force $F_g$ itself. The second observation tells him that a constant (which we will refer to as Kepler's constant) $K$ will probably play a role in the law of gravitation that he is seeking. This constant is the product of the square of the angular frequency $\omega$ with the cube of the radius $r$ of the planetary orbital motion: $K = \omega^2 r^3$. And finally, apart from $g$ and $K$, the distance $r$ is expected to feature in the equation sought.
So, Newton was seeking a relation $f(K, r, g) = 0$. Kepler's constant $K = (2 \pi)^2 AU^3 yr^{-2}$ has dimensions $[L^3 T^{-2}]$, the distance $r$ has dimension $[L]$, and the acceleration $g$ has dimension $[L T^{-2}]$. So one dimensionless parameter can be defined: $r^2 g/K$. This directs Newton towards an equation of the form $g \propto K/r^2$. In terms of the force $F_g = m \ g$, this reads:
$$F_g \propto K \frac{m}{r^2}$$
However, according to Newton's third law, this force needs to be mutual between the two bodies. If $M$ denotes the sun's mass, it follows that $K = G M$ with $G$ a universal constant equal to $(2 \pi)^2 M_{sol}^{-1} AU^3 yr^{-2}$. The result is an equation symmetric in both masses:
$$F_g \propto G \frac{M \ m}{r^2}$$
Obviously, we can replace the proportionality sign with an equal sign provided we absorb any mathematical factors hidden from the dimensional analysis into the constant $G$.
You've made a mistake in saying
Here, we have $n = 3$ variables, and one may say that we also have $k = 3$ fundamental dimensions $M$, $L$ and $T$.
but it's a subtle one.
To see why the implication does not hold, consider three different variables, with dimensions $[A]=ML/T$, $[B]=ML/T$ and $[C]=ML/T$. Here you have three variables, and three independent dimensions $M$, $L$ and $T$... but of course you cannot infer from their dimensions alone that $A$, $B$ and $C$ are themselves independent. In my example, $n=3$, $k=1$, I can form $n-k=2$ independent pi groups, $B/A$ and $C/A$, and any physically meaningful equation $f(A,B,C)=0$ can be rephrased as $F(B/A,C/A)=0$. You can see how it would go if you add more variables $D,E,\ldots$ with the same dimensionality.
(Note that this is quite a common situation. For example, $B$ and $C$ might be the momentum of two particles which are joined in a collision to give a total momentum $A$. The correct relationship between them is then $F(b,c)=b+c-1=0$.)
The way to test for this is via the dimensionality matrix. In your case, with $[F]=ML/T^2$, $[\eta]=M/T$ and $[v]=L/T$, the dimensionality matrix is
$$
D=\begin{array}{cl}
\begin{array}{ccc}F&\eta&v\end{array} & \\
\left(\begin{array}{ccc}1 & 1 & 0 \\ 1 & 0 & 1 \\ -2 & -1 & -1 \end{array}\right).
&\begin{array}{c} M\\ L\\ T \end{array}
\end{array}
$$
Its rank is $k=2$, which means that you really only have two independent dimensionalities. Any two will do, but the clearest choice is probably $[\eta]$ and $[v]$, in which case $[F]=[\eta][v]$ is obviously dependent. Since there's $k=2$ independent dimensions and $n=3$ variables, you can form $n-k=1$, i.e. a single pi group, which is $F/\eta v$ or its inverse. Your initial physically-meaningful equation $g(F,\eta,v)=F-\eta v=0$ can thus be rephrased as $$G(F/\eta v)=F/\eta v-1=0.$$
Note in particular that the pi groups need to be dimensionless. This is the whole point of the pi theorem and the game doesn't really make sense unless you do that. Thus
Skipping the work, $\Pi_0$ will correspond to $F = \eta\dot{x}$.
is not correct.
Similarly,
cases where $j=n-k<0$
don't make sense. You cannot encompass $k>n$ independent dimensions using only $n$ variables. This is a fundamental fact of linear algebra: a set of $n$ vectors (in dimensionality space) can only span a subspace of dimension $k\leq n$.
It may be worthwhile exploring the other end of the dimensionality scale to clarify what's going on - the case where $n=k$. Here all your dimensions are physically independent - you might have a length, a time and a mass, say. There is then nothing you can say about them that will be physically meaningful, and you can't form a single pi group. The only allowed statement is $F()\equiv0=0$, which is an empty statement. You can define new variables (like, say, momentum) but until you do you're stuck with nothing.
Best Answer
There is no answer to this. When you are taught to use dimensional analysis at school the teacher invariably selects an easy example (it's almost always the pendulum) to keep things simple. In the real world there is no guarantee that you have a dimensionless constant.
It's actually quite rare to use dimensional analysis to derive equations in the real world. The sorts of simple systems that are amenable to dimensional analysis are usually already well known. However it's very, very useful to use dimensional analysis to check that an equation you derive is dimensionally consistent.
For example suppose you're working through a differential equation for some quantity, and after covering many sheets of paper with scribbles you end up with a final equation. It's very easy to make a minor mistake along the way, so the first thing you check is that your final equation is dimensionally consistent, i.e. the dimensions of the left and right sides are the same. If they aren't that means you've made a mistake somewhere. I routinely do this in my answers to questions on this site!