Can an amide nitrogen be a hydrogen bond acceptor?
If I understood you correctly, you are talking about the peptide bond nitrogens ($\ce{R-C(=O)-\color{red}{N}H-R}$). This is, reduced to its significant chemical functional group an amide, more precisely a carboxylic amide.
The amide nitrogen technically has a lone pair and thus technically could function as a hydrogen bond acceptor when viewed alone. However, this lone pair is actually significantly delocalised towards the carbonyl bond. If we ignore the two $\ce{R}$ residues, we could write this the following way:
$$\ce{O=C-N-H &lt;-&gt; ^{-}O-C=N^{+}-H}$$
The lone pair is not really available as a lone pair on the nitrogen, but forms a π-bond to the neighbouring carbonyl carbon to a significant extent. Thus, these nitrogens cannot be hydrogen bond acceptors.
Further, the fact that a Grignard reagent cannot react with an amide gives great support for the fact that nitrogen’s electrons are delocalized onto the carboxylate.

Finally, thank you Orthocresol for supplying me with the paper given at the end of this post. The authors performed some quantum chemical calculations to extract the energy difference between free formamide and water molecules and various conceivable hydrogen bonds between them. The strongest stabilisation can be found in an amide–amide hydrogen bond of the following type:
$$\ce{H2N-CH=O\bond{...}H-NH-CO-H}$$
At around $-40~\mathrm{kJ \cdot mol^{-1}}$ for a $\ce{H\bond{...}O}$ distance of $285~\mathrm{pm}$ and a $\ce{C=O\bond{...}H}$ angle of $60^\circ$.
$$\ce{H2N-CH=O\bond{...}H-OH}$$
For the interaction of a water molecule donor and the carbonyl oxygen as depicted just above, the stabilisation is given as up to $-27~\mathrm{kJ \cdot mol^{-1}}$ for a distance of $280~\mathrm{pm}$ and a $60^\circ$ $\ce{C=O\bond{...}H}$ angle.
$$\ce{OHC-HN-H\bond{...}OH2}$$
For this interaction where the amide nitrogen is donor and water is acceptor, the stabilisation was calculated to $\Delta E \approx - 30~\mathrm{kJ \cdot mol^{-1}}$ — achieved twice. Once for the geometry $d(\ce{H\bond{...}O}) = 280~\mathrm{pm}$ and $\angle(\ce{H\bond{...}O-H2}) = 30^\circ$ (the angle is measured between the $\ce{H-O-H}$ plane and the amide plane), and once for $d = 280~\mathrm{pm}$ and $\angle = 15^\circ$.
$$\ce{OCH-H2N\bond{...}H-OH}$$
This is the one you are looking for. These energy differences are all given as positive. The lowest is $\Delta E \approx +0.8~\mathrm{kJ \cdot mol^{-1}}$ for a distance $d(\ce{H\bond{...}N}) = 360~\mathrm{pm}$ — $1.3$ times the previous distances. (The angle is not given but from the picture we can assume a $90^\circ$ angle between the amide’s plane and the $\ce{N\bond{...}H}$ axis.) For the $280~\mathrm{pm}$ which brought the most stabilisation in the other cases and which thus seems to be the ideal hydrogen bond distance, they report $\Delta E \approx + 30~\mathrm{kJ \cdot mol^{-1}}$ — this is about the dimension of the stabilisation of the previous hydrogen bond only in a destabilising manner!
These studies let us conclude that it is safe to say the amide nitrogen will never be a hydrogen bond acceptor.
Reference:
A. Johansson, P. Kollman, S. Rothenberg, J. McKelvey J. Am. Chem. Soc. 1974, 96, 3794. DOI: 10.1021/ja00819a013.