Yes the product $\nu \lambda$ makes sense as a velocity. Defining $E = \hslash \omega$ and $p=\hslash k$ (the Planck constant $h=2\pi \hslash$, where the $2\pi$ is injected into the $\hslash$, since physicists usually prefer to discuss the angular frequency $\omega=2\pi\nu$ and the wave vector $k=2\pi p$ rather than the frequency $\nu$ and the momentum $p$ for Fourier-transform notational convenience), you end up with
$$\nu \lambda = \frac{\omega}{2\pi}\frac{2\pi}{k}=\frac{\omega}{k}$$
which is the standard definition for the phase velocity. It corresponds to the velocity of the wave component at frequency $\nu$ propagating over distance $\lambda$ per unit time.
This is always true, but for more complicated situation, a system is represented by a superposition of different waves propagating at different phase velocity. It results a wave-packet propagating, as an ensemble, at the group velocity
$$v_{g}=\frac{\partial \omega \left(k\right)}{\partial k}$$
where the dispersion relation of the wave-packet is noted $\omega \left(k\right)$.
Only the group velocity has some physical clear interpretations. For instance, the phase velocity can be larger than the speed of light, but the group velocity can never been larger than the speed of light $v_{g}\leq c$, at least in vacuum.
For a photon in free space for instance, $\omega \left(k\right)=c k$ and thus it's group velocity is $c$. For a free non-relativistic particle of mass $m$, $\omega = \hslash k^2/2m$, and $v_{g}=\hslash k/m$. etc...
For a human body, you have to count all the atoms constituting the body. The individual frequency of one atom interferes with the frequencies of all the others, resulting in an almost flat dispersion relation. The group velocity is then ridiculously small. A human body does not move due to quantum effect ! You can get a basic idea of the group velocity of a human body supposing you are a free particle of mass $100 \textrm{kg}$ and wavelength $1 \textrm{m}^{-1}$ (i.e. the order of magnitude of your size is about $1 \textrm{m}$) the $\hslash$ kills $v_{g} \sim \left(10^{-36}-10^{-34}\right) \textrm{m.s}^{-1}$ !
More about that :
Thinking about this answer:
$$KE(rel)=(\gamma-1)mc^2$$
Thus, if T=KE(rel):
$$\gamma=1+T/mc^2$$
If $T>>2mc^2>mc^2$:
$$\gamma\approx T/mc^2$$
By the other hand, from the de Broglie fundamental relationship:
$$\lambda=\dfrac{h}{p}=\dfrac{hc}{pc}=\dfrac{hc}{m\gamma v c}$$
and then, with $\gamma\approx T/mc^2$, and $v\approx c$ we obtain
$$\lambda=\dfrac{h mc^3}{m T v c}=\dfrac{h c}{T}$$
From this last equation we get the above one from a simple use of the dispersion relationship. However, as the kinetic energy is much bigger than the rest mass, my doubt concerning the presence of mass remained...Until I did some numbers...
Best Answer
$p=mc$ is NEVER true.
Like you mentioned, for a particle with relativistic velocity (v is not negligible compared to $c$), the right equation is:
$$p = \gamma m v = v\cdot \frac{m}{\sqrt{1-(\frac{v}{c})^2}} $$ Now if you try to apply it to photons, $ m=0$ and $\sqrt{1-(\frac{v}{c})^2}=\sqrt{1-1}=0$ so you get a $\frac{0}{0}$ situation - which gives us no information at all! So you can't apply this formula for photons.
But, there is another, more accurate equation:
$$E^2 = (pc)^2 + (mc^2)^2 $$
So here, for photons, $m=0$ and we get $$E^2=(pc)^2 \Rightarrow E=pc$$ Side note: that was known before Einstein, you can get this result straight from Maxwell's equations and the Lorentz force.
Now, if $E=pc$, then $p=\frac{E}{c}$ , which for a single photon means $$ p = \frac{E}{c} = \frac{h\frac{c}{\lambda}}{c} = \frac{h}{\lambda} $$
I hope it helped :)