Difference in mechanisms between formation of acetal and enamine When forming acetal from a C=O group with alcohol, the alcohol will act as a nucleophile and attack twice as the carbon is made very electrophilic twice.
I wonder in the last intermediate before forming the enamine, is it able for the 2' amine to act as a nucleophile and attack the carbon bonded to a highly electronegative N with a positive formal charge? (like the attack from the alcohol when forming acetal) Then with the help with 1 proton transfer, forming a product that has 2 RR'N?
My thought is that it may be too sterically hindered compared to the alcohol when attacking. But is it valid?
This does indeed work. One such reference includes US patent 3696080A in which propionaldehyde is reacted with dimethylamine to give the N,N-acetal as shown in the figure below.

The described reaction procedure involves cooling both reagents to $\pu{-78^\circ C}$ before addition, letting the temperature rise to around $\pu{-35^\circ C}$ at which point anhydrous potassium carbonate was added, letting the temperature rise again to about $\pu{0^\circ C}$ at which point the cooling apparatus was switched from dry ice/acetone to ice/water, and then stirring it overnight allowing the ice to melt in the process. The workup the inventors describe is again rather complex but they claim to isolate up to $\pu{73\%}$ of the bis-amino product.
SciFinder also gave a number of other references but one should serve sufficiently as a proof of concept.
It’s worth noting that $\pu{2.4 eq}$ of amine were added which probably assisted in driving the reaction to the product side.