Obtaining values for electrical mobility from ion transport numbers in a Hittorf cell
You can't directly derive electrical mobilities from ions from their transport numbers; transport numbers are a relative measurement of charge transfer that, precisely, reduces absolute mobilities for different ions to contributing fractions. Absolute mobilities are lost when we use transport numbers, so if all we know about an electrochemical system is transport numbers, we cannot recover ionic mobilities.

However, we can recover drift velocities if we have additional information on the system - specifically, total current $I_\mathrm{total}$ and distance between plates in the cell $d$. Alternatively, if we're interested in ion mobilities, we will also need to know voltage $V$.

Let's look at a simple system with only one electrolyte involved in the electrochemical reaction. Total current will be the sum of the contributions of the anion and the cation:

$$ I_\mathrm{total} = I_+ + I_- =  \cfrac{n_+ z_+ \nu_+ e}{d} +  \cfrac{n_- z_- \nu_- e}{d} $$

where $n_i$ is the number of ions $i$, $z_i$ is the charge of ion $i$, $\nu_i$ is the drift velocity of ion $i$, $e$ is the electron charge and $d$ is the distance between the plates. Per the requirement of electroneutrality,

$$ n_+ z_+ = n_- z_- $$

so

$$ I_\mathrm{total} = \cfrac{ n_+ z_+ e \left( \nu_+ + \nu_- \right) }{d} = \cfrac{ n_- z_- e \left( \nu_+ + \nu_- \right) }{d} \tag{1}$$

We can calculate transport numbers as

$$ t_+ = \cfrac{I_+}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_+ e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_+}{\nu_+ + \nu_-} $$

$$ t_- = \cfrac{I_-}{I_\mathrm{total}} = \cfrac{n_+ z_+ \nu_- e /d}{n_+ z_+ \left( \nu_+ + \nu_- \right) e / d} = \cfrac{\nu_-}{\nu_+ + \nu_-} $$

Note that we could use the definition of transport numbers to obtain a specific drift velocity, $\nu_i$, from the total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ = t_+ \left( \nu_+ + \nu_- \right) $$

$$ \nu_- = t_- \left( \nu_+ + \nu_- \right) $$

Let's go back to $\mathrm{(1)}$ to derive an expression for total drift velocity, $\nu_+ + \nu_-$:

$$ \nu_+ + \nu_- = \cfrac{d}{n_+ z_+ e} I_\mathrm{total} = \cfrac{d}{n_- z_- e} I_\mathrm{total} $$

We can therefore derive drift velocity for an ion, $\nu_i$, from its transport number, $t_i$:

$$ \nu_+ = \cfrac{ t_+ d }{ n_+ z_+ e } I_\mathrm{total} $$

$$ \nu_- = \cfrac{ t_- d }{ n_- z_- e } I_\mathrm{total} $$

Note that it is necessary to know current $I_\mathrm{total}$ and cell size $d$ since precisely these values introduce the time and distance dependence necessary to recover drift velocity $\nu_i$ from transport number $t_i$.

Assuming the system can be approximated by parallel plates, i.e. assuming electric field is constant, ion mobility, $\mu_i$ can be derived from drift velocity through voltage, $V$, and distance between plates, $d$:

$$ \mu_i = \cfrac{\nu_i}{E} = \cfrac{\nu_i d}{V} $$

So the expressions for cation and anion mobility would be

$$ \mu_+ = \cfrac{t_+ d^2}{n_+ z_+ e V} I_\mathrm{total} $$

$$ \mu_- = \cfrac{t_- d^2}{n_- z_- e V} I_\mathrm{total} $$

For a more complex system with $N$ ions contributing to current,

$$ I_\mathrm{total} = \cfrac{e \sum\limits^{N} n_i z_i \nu_i }{d} $$

$$ t_i = \cfrac{ n_i z_i \nu_i }{\sum^{N} n_j z_j \nu_j } $$

and drift velocity and mobility can be computed as 

$$ \nu_i = \cfrac{ t_i d }{ n_i z_i e} I_\mathrm{total} $$

$$ \mu_i = \cfrac{ t_i d^2 }{ n_i z_i e V} I_\mathrm{total} $$