The Magnus effect was discovered when an explanation for the low precision of guns was needed. It affects the cylindrical, pointed grenades just as much as any ball.
It does not matter how long the rotating body is: Once it rotates, it will create a low pressure area on one side orthogonal to the crosswind direction and a corresponding high pressure area opposite to it.
If the body rotates around its short axis, friction will slow the rotation down much more quickly compared to a spherical body. This could be the main difference between a rotating rugby ball and a rotating soccer ball.
No, you are mixing up concepts at random. First of all, there's no reason for pressure to increase just because the surface area of the tube decreases. The definition of pressure $p=F/A$ does not allow you to draw this conclusion, since you have no way (without additional information) to know what happens to the numerator of this expression.
Second, you need to keep your boundary conditions straight: Are you performing an experiment where you keep the flow rate constant? If that is the case then, yes, your pressure gradient will increase with decreasing tube radius, and the velocity will increase, of course. Often, however, we keep the pressure gradient constant (e.g. in a gravity-driven flow), in which case your flow rate decreases with the tube radius. Or, you may have a pump driving your system that has constant available power, in which case the product of pressure drop times flow rate is constant, and reducing the tube diameter may both increase the pressure gradient and decrease the flow rate, depending on the type of pump you use. In this case the flow velocities may or may not increase. Long story short, no, you're wrong on this one, too.
Finally, no, your ideas around how and why the pressure changes with reduced tube diameter and the connection with particle collisions with the wall are wrong. The fact that the Bernoulli equation assumes frictionless flow has been mentioned already.
Best Answer
This problem can get as complicated as you like, but let me show you how it can be approached in the most simple case. The Navier-Stokes equations describe the flow:
$\rho\left(\displaystyle\frac{\partial {\bf u}}{\partial t}+({\bf u}\cdot\nabla){\bf u}\right)=-\nabla p + \mu \nabla^2 {\bf u}$. $\qquad$ (1)
Assume that the motion of the fluid is slow enough so that the non-linear term $({\bf u}\cdot \nabla){\bf u}$ is negligible. This will be true as long as viscosity forces are large compared to inertial forces from acceleration. Ugh, now that it is a linear equation, something can be done analytically.
Consider the problem in cylindrical coordinates with $z$, $r$ and $\phi$. Let us assume $\partial/\partial z=0$ and $\partial/\partial \phi=0$ for all quantities due to the axial symmetry of the problem. Then the continuity equation will offer additional simplifications. In cylindrical coordinates, the continuity equation reads
$\displaystyle\frac{\partial \rho}{\partial t}+ \frac{1}{r}\displaystyle\frac{\partial(\rho r u_r)}{\partial r}+ \frac{1}{r}\frac{\partial\rho u_{\phi}}{\partial \phi} + \frac{\partial (\rho u_z)}{\partial z}=0$
With $\rho=\mathrm{const}$ (incompressibility assumption), the first term is zero, and last two terms are equal to zero due to the axial symmetry assumption. The remainder yields $u_r=0$. Hooray! Let us choose the problem so that $u_z=0$, too (true if the cylinder is not moving with respect to the fluid in $z$-direction) Then we are left with just $u_{\phi}$, and let us denote it $u\equiv u_{\phi}$.
If the cylinder is suspended vertically, i.e., the axis of the cylinder is along the gravity force, then the pressure gradient $\nabla p$ can also be neglected. This is because the only component of $\nabla p$ is in z-direction due to gravity, and we already assumed $\partial {\bf u}/\partial z$=0. And there may be no $\phi$-component of $\nabla p$, because it would violate our $\partial /{\partial \phi}=0$ assumption.
All of the above leaves us with the only equation for $u_\phi\equiv u$:
$\displaystyle\frac{\partial u}{\partial t} = \mu \frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial u}{\partial r}\right)$. $\qquad$ (2)
To understand where it came from, look up the explicit form of equation (1) in cylindrical coordinates at this Wikipedia page, take the $\phi$-component of it and use the assumptions outlined below to get rid of $\partial/\partial z$ and $\partial/\partial \phi$ terms, and of the non-linear and pressure terms.
The cylinder is spinning, which makes the boundary conditions for equation (2)
$u(r=0, t)=0$ and $u(r=R, t)=R\omega_0$,
where $R$ is the radius and $\omega_0$ is the angular frequency of the cylinder. To understand the second one, look at the well-known solution of a similar problem at this Wikipedia page or this paper referenced in the Wiki page. The initial conditions are
$u(r, t=0)=0$.
Equation (2) is a very well-known equation. It is the diffusion or the heat equation in disguise. Well, no surprise about it, because viscosity in is nothing more than diffusion of momentum. At least, this is how it is expressed in the viscosity term of equation (1). The solution is readily obtained using separation of variables and Fourier analysis. I will not solve it here, because it is a well understood mathematical problem, with which graduate students are routinely tortured with great success. This Wikipedia page describes it.
And the back-of-the envelope estimate I mentioned in my comment comes from dimensional analysis of equation (2). If the typical velocity scale of the problem is $U$, spatial scale is $R$ and time scale is $T$, then (2) can be approximated as
$\displaystyle\frac{U}{T} \approx \mu \frac{1}{R} \frac{1}{R} R \frac{U}{R}$,
or
$T \approx R^2/\mu$,
which is a crude estimate of your spin-up time.