What is the maximum (unloaded) range of the Beluga XL?
For an estimate, we can use the Brequet range equation:

$range = V\cdot t_f 
= \frac{L}{D} \times I_{sp} \times \ln\left(\frac{W_i}{W_f}\right) $ 

with:


$V$: optimal velocity
$t_f$: flight time
$L$: lift
$D$: drag
$I_{sp}$:  propulsion efficiency
$W_i$: initial weight
$W_f$: final weight


Assuming the lift to drag ratio $\left(\frac{L}{D}\right)$ and the propulsion efficiency will stay constant, the range will depend on the ratio of initial and final weight. 

$range = C \times \ln\left(\frac{W_i}{W_f}\right)$

In case the aircraft departs with the maximum payload, the weight is:

$W_i = W_{oe} + W_{payload} + W_{fuel} = W_{mto}$


$W_{oe}$: operating empty weight: 127 500 kgf
$W_{payload}$:  payload weight: 51 000 kgf
$W_{mto}$: maximum take-off weight: 227 000 kgf
$W_{fuel}$: fuel weight =  $W_{mto} - W_{payload}-W_{oe}$ = 48 500 kgf


The final weight at maximum range is then: 
$W_f = W_{oe} + W_{payload}$

Filling in the numbers we find:

4000 km = $C \times \ln\left(\frac{\textrm{227 000}}{\textrm{127 500 + 51 000}}\right)$

That means $C \approx \textrm{16 640}$ for the Beluga XL.

If we now remove the payload and take-off with the same amount of fuel, we find:


$W_i$ = 176 000 kgf
$W_f$ = 127 500 kgf
$range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{176 000}}{\textrm{127 500}}\right) \approx 5365$km


But that is just removing the payload without adding fuel. Of course we can now bring more fuel on board without exceeding the maximum take-off weight. 

The fuel capacity of the Beluga XL is 73000 kg (source:Aircraft Characteristics
Airport And Maintenance Planning Beluga XL manual). This changes the numbers to:


$W_i$ = 200 500 kgf
$W_f$ = 127 500 kgf
$range \approx \textrm{16 640} \times \ln\left(\frac{\textrm{200 500}}{\textrm{127 500}}\right) \approx 7530$km


So the Beluga can fly approximately 7500 km when empty, enough to cross the Atlantic from Toulouse, even if there is normal headwind.