How to calculate actual hfe for low voltage source?
Any design that requires the exact value of the \$h_{FE}\$ of a BJT is a faulty design, since that parameter is subject to wild production spread.

Since in your case you only need the \$h_{FE}\$ to calculate the base current required to turn the BJT on, assume the worst possible case reported in the datasheet.

In this case worst possible case means lowest possible value of \$h_{FE}\$ in your working conditions. For any small signal BJT assuming \$h_{FE}\$ to be at least 10 (for the purpose of turning it on) is almost sure-fire. In other words, use a base current that is ten times smaller than the maximum collector current you need to drive and you'll be safe. 

If the current you must provide to the base turns out to be too big for your driver circuit, you can use a Darlington transistor (or another BJT connected as a Darlington pair). In which case the minimum \$h_{FE}\$ (without even looking at the datasheet) can be assumed as 100 (VERY conservative estimate for a Darlington BJT, it is likely higher; it is OK for two 2N2222 connected as a Darlington pair).

If you want to be more fussy, the \$h_{FE}\$ depends essentially on the \$I_C\$ an temperature. If you want a tighter estimation take the lowest value you find in the datasheet, as you can see in an excerpt of a Fairchild PN2222 datasheet (a modern e cheaper equivalent of the now obsolete 2N2222):



In this case it is generally safe to assume a min \$h_{FE}\$ value of 35.

EDIT (prompted by a comment)

Computing the \$h_{FE}\$ when the BJT is saturated is not meaningful. \$h_{FE}\$ is a parameter which has some meaning when the BJT operates as a linear device, i.e. as an amplifier. When you turn the BJT on you are actually overdriving the BJT out of its linear region (a.k.a. active region).

When you perform your calculations to determine how to bias the BJT to get the level of overdrive needed to turn it on, you are simply using \$h_{FE}\$ as a gross indicator of where the device will stop behaving linearly. In other words you drive it with a base current that cannot be amplified linearly by the device, so it is forced into saturation (i.e. it is turned on).

To be more explicit, let's say that the min \$h_{FE}\$ is 10. This means that if the device is driven with 1mA Ib it will have a 10mA Ic if the output circuit allows it. If, for a given level of Ic, say 10mA, you drive the base with much more than Ic/10=1mA, you are forcing the BJT out of the linear region (where "there is no \$h_{FE}\$").