With what velocity are we moving along the time dimension? Does the question make sense? Velocity along time axis means $v_t=\mathrm dt/\mathrm dt$? If it doesn't, please explain where the flaw is. Taking time as measure like length? Or do we need to differentiate time with respect to some other quantity?
Extension of the question is welcomed.
 A: At exactly 1 second per second.
A: In relativity the time coordinate is $x_o=ct$ and its time derivative (in the rest frame) is $c$. Therefore the time component of four-velocity is the speed of light in vacuum.
A: In non-relativistic mechanics, time $t$ is a (universal) parameter and a particle's coordinates (in some inertial coordinate system) can be expressed as three functions, $x(t),y(t),z(t)$ of this universal parameter $t$.  The velocity of the particle (in these coordinates) is then the derivative of the position with respect to the parameter $t$:
$$\mathbf{v} = \frac{dx}{dt}\hat{\mathbf{x}} + \frac{dy}{dt}\hat{\mathbf{y}} + \frac{dz}{dt}\hat{\mathbf{z}}$$ 
However, in relativistic mechanics (SR for simplicity), time $t$ is a coordinate which is reference frame dependent.  Still, the world line of a particle can be parameterized with the proper time $\tau$ which is essentially the time of an ideal clock fixed to the particle ('wristwatch time').
The coordinates of the particle (in some inertial coordinate system) can then be expressed as four functions, $t(\tau),x(\tau),y(\tau),z(\tau)$ of the particle's proper time $\tau$.  The four-velocity of the particle is then the derivative of the four-position with respect to the parameter $\tau$:
$$\vec{U} = c\frac{dt}{d\tau}\hat{\mathbf{t}} + \frac{dx}{d\tau}\hat{\mathbf{x}} + \frac{dy}{d\tau}\hat{\mathbf{y}} + \frac{dz}{d\tau}\hat{\mathbf{z}}$$
So, in this coordinate system, the component of the particle's four-velocity in the time direction is 
$$U^0 = c\frac{dt}{d\tau}$$
Now, it can be shown that (time dilation)
$$dt = \gamma_v d\tau$$
where
$$\gamma_v \equiv \left(1 - \frac{v^2}{c^2}\right)^{-1/2}$$
and
$$ v = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2}$$
thus
$$U^0 = c\gamma_v$$
This is, I believe, a reasonable answer to the question "With what velocity are we moving along the time dimension?" if, by velocity, one means the the derivative of the coordinates with respect to a time parameter.
(note:  as I was finishing typing this answer up, I noticed that Ben Crowell had posted essentially the same answer but I'll post this anyhow since it's already done.)
A: There is a variety of different conventions for defining some of the details, but the most common way to describe this, among relativists, would be the following. We take units in which $c=1$. There is a velocity four-vector which is tangent to the world-line of a particle. The normalization of this four-vector is defined so that its norm is 1 (in $+---$ signature). All of this is coordinate-independent.
If we now specialize to Minkowski coordinates $(t,x,y,z)$ in flat spacetime, then the components of the velocity four-vector become the derivative of the coordinates with respect to proper time $\tau$ (not coordinate time $t$), and the normalization condition ends up causing the timelike component of the velocity vector to be the Lorentz factor $\gamma$. This is the closest thing we have, in common professional notation, to a useful way of defining something that is useful and corresponds in some way to the notion of a "velocity along the time dimension." It's $\gamma$.
In the special case where the particle is at rest with respect to the Minkowski frame being used, we have $\gamma=1$. This is the justification you see for the statement in popularizations that we "move through spacetime at the speed of light," since the speed of light is 1. However, most relativists cringe at this phraseology, which seems to have been propagated by Brian Greene.
