Here's my quantitative attempt at $4.$ and $1.$:
The Coandă effect here is the tendency of the airflow to adhere to the surface of the ball. This means that near the surface of the ball, the streamlines are curved with a radius of curvature approximately equal to the radius of the ball $R$; this curvature results in a pressure gradient just as it does in the theory of lift which I assume to be constant over the surface:
$\vec{\nabla} P = - \frac{\rho v^2}{R} \vec{e}_r$
(Anecdotally, the "circulation" aerodynamicists use to calculate the lift per unit width is related to a certain integral of this quantity over an appropriate cross sectional area.)
This points radially inwards to some equilibrium point, at which the net forces acting on the ball cancel. In practice, the ball will hang slightly below this point due to gravity, Edit: see below.
When the ball is displaced from this position normally to the flow by some distance $z$, we can calculate the total restoring force acting on the ball by integrating $\vec{\nabla} P$ over the volume of the ball under two assumptions:
$1.$ the curvature and equilibrium position of the airflow is negligibly disturbed by moving the ball like this, and so independent of $z$
$2.$ the net volume of the displaced ball that's relevant to the integration (i.e. that contributes to an *un*balanced force) is approximately equal to $\pi R^2 z$, the cross sectional area of the ball times the displacement $z$
with these assumptions, the restoring force is
$\vec{F} = - \frac{\rho v^2}{R} \pi R^2 \vec{z} = -\pi \rho v^2 R \vec{z}$
which we can immediately recognise as a harmonic oscillator
$\vec{F} = -k \vec{z} = -m \omega^2 \vec{z}$
where $m$ is the ball's mass, and get for natural frequency
$\omega = v \sqrt{\frac{\pi\rho R}{m}}$.
or for an easier-to-measure frequency of oscillation $\nu$ in terms of $Q$, estimating $v = Q/A$,
$\nu = \frac{Q}{A} \sqrt{\frac{\rho R}{4 \pi m}}$
using the cross sectional area of the ball for $A = \pi R^2$, because $v$ is the velocity at which air travels over the ball's surface, and not the velocity at which it leaves the pen.
This assumption is a bit questionable as none of the air actually passes through this area, but this is the only finite area relevant to the problem. At worst, the true area that this depends on, probably the "average" area that the flow passes through in the plane that passes through the ball's equator, will be $A$ times some numerical constant.
It's important to note that this scales linearly with $Q$, so blowing twice as hard will result in oscillations that are twice as fast.
The international standards are $R$ = 20 mm and $m$ = 2.7 g, so I would guess the following formula:
$\nu = 0.767$ m$^{-1} \frac{Q}{A}$
Using BebopButUnsteady's reasonable $Q$ of $6$ L/s, and the above $R$,
$\nu = 3.66$ Hz
Which sounds fairly reasonable to me. But again, the linear dependence on $Q$ is important, and for weaker flows it will be proportionately slower.
And as a final note, my earliest assumption that the pressure gradient was constant over the surface means this formula doesn't distinguish between oscillations parallel the stream and oscillations normal to it.
In practice, it's known that the inward pressure gradient will be lower at the points of the ball that are "in line with" the stream than those that lie further outside since the streamlines curve in the opposite direction, so oscillations parallel to the stream will have a lower frequency than the normal/"lateral" ones in general.
PS: The final formulas here will probably only hold up to some experimental "fudge factor", which may be interpreted as the numerical constant that appears in that approximation of $v \approx Q/A$: if $v = f Q/A$, then $f$ is the fudge factor. But this is fluid mechanics, and fudge factors are inevitable.
Edit: estimate for part $1.$
If we assume that the ball's displacement due to gravity is less than one radius, then we can say that
$$mg < \pi \rho v^2 R^2 = \rho Q^2 / \pi R^2$$
given the above definitions of $R$ and $Q$, or
$Q > R \sqrt{\frac{\pi mg}{\rho}} \approx 5.77$ mL/s
and
$v > \frac{1}{R} \sqrt{\frac{mg}{\pi \rho}} \approx 4.59$ m/s
which is far less than $6$ L/s, but may still be in line with what those toys that sometimes come in Christmas crackers. The velocity looks somewhat more intuitively reasonable. If these are too low to sustain experimentally, the reason for this may be that the flow is too unsteady when maintained by human breath.
In some videos, the amplitude of oscillations when suspended in a hair-drier flow can apparently be appreciably larger than one radius, so the assumption about $z<R$ may be out by some factor of up to 5; however, assuming this would make the final results for theoretical minimum $v$ or $Q$ smaller by a factor of $\approx 0.44$, so it is still an informed guess for an upper bound.
Since the assumptions involved in deriving this force only strictly apply to displacements normal to the stream, this is the minimum flow to suspend the ball horizontally; answering the question about maximum suspension angle (there may not be one) would require a more detailed description of the pressure distribution around the ball, which may well be available in the literature for laminar flow around a sphere.
Best Answer
I've confirmed the experiment, using a McD_n_lds paper drinks cup and a beer can hollow plastic ball of about $5\mathrm{g}$, of about the same diameter as a ping pong ball (PPB):
The observed effect depends largely on the cup being soft and permanently deformable (like an object made of blutack or playdough), so its collision with Earth is inelastic. A stiff, hard cup (made of steel e.g.) would not work the same way here. The inelastic collision of the ensemble causes kinetic energy of cup and water, post-collision, to be small.
The PPB bounces back quite high (from a quarter-filled cup) and the cup of water loses quite little water and doesn't really bounce at all. It's quite a sight to behold! A simple model can be set up a follows.
We can write with Conservation of Energy (the collision is clearly not elastic - as evidenced by the permanent deformation of the bottom of the cup):
$$(M+m)gH=mgh+W+\Delta Q+K_{M+m}$$
where:
Trouble is, we don't know the value of $W+\Delta Q+K_{M+m}$. Direct observation suggests it is small, so we can write:
$$(M+m)gH\geq mgh$$
Or:
$$\boxed{h \leq H\Big(\frac{M+m}{m}\Big)}$$
If $M\gg m$ we can further approximate:
$$h \leq \frac{M}{m}H$$
I wanted to confirm experimentally the effect of $M$ on $h$.
Using a nearly empty cup, one half-filled and one filled completely I can confirm increased $M$ increases $h$.
Some further experiments are planned.