The Concentric Bordering  

Augmentation factor Δ_{p,q} = (p^{n}q^{n})/2 
the augmentation factor of the order q subhypercube  
The center of the order m hypercube is (m^{n}+1)/2 so Δ_{p,q} = (p^{n}+1)/2  (q^{n}+1)/2 = (p^{n}q^{n})/2 

Augmented sum ΔS(H^{[(pq)]}_{q}) = q(p^{n}q^{n})/2 
the augmentation of the sum of the order q subhypercube  
Each cell of the subhypercube is augmented by (Δ_{p,q}) which therefore is counted q times. 

Concentric border sum S(B_{p,q}) = (pq)(p^{n}+1)/2 
Since the order q subhypercube is an augmented magic hypercube the borders 1agonal and nagonal numbers need to sum up to the same constant. 

The remaining sum the to be corrected by the border is the difference in the magic sums of the target order p and the suborder q so: S(B_{p,q}) = S(H_{p})S(H^{[(pq)]}_{q}) = p(p^{n}+1)/2  q(q^{n}+1)/2  q(p^{n}q^{n})/2 = (pq)(p^{n}+1)/2 

The BorderFlip [0,0] <=> [m1,m1] [0,m1] <=> [m1,0] [i,0] <=> [i,m1] i=1..m2 [0,i] <=> [m1,i] i=1..m2 
A "borderflipped" magic square is still a magic square  
With the exchange of the numbers in the mentioned positions all 1agonal and 2agonals remain having the same numbers. 

Augmenting square [Collapsed Latin square] * (m^{n}m^{2})/(m1) (square factor: m) 
A square augmenting the number from the square to the hypercube range  
A Latin square can be collapsed by a digit change with duplicated digits as long as the sum of those digits remain m(m1)/2 the squares sum remains multiplying this collapsed square with [m(m^{n}+1)/2m(m^{2}+1)/2]/[m(m1)/2] = (m^{n}m^{2})/(m1) is sufficient add onn to augment the sums as appropriate. for the square this reduces to [m(m^{2}+1)/2m(m+1)/2]/[m(m1)/2] = m 

The above outlines the basic principles, the sample in the table below show that imperfections in the augmenting hypercube can be corected in the augmented hypercube (current expertise suggest these to be forced upon this scheme due to "natural overlap") 

Due to lack of expertise the following are some inexact qualitative statements Future upload these statements needs to be exactified. 

"natural overlap"  reason for the imperfection in the borders  
The number range of the bordered hypercube is
[(p^{n}+1)/2(q^{n}1)/2,[(p^{n}+1)/2+(q^{n}1)/2] Not clear yet how to formulate this phenomonon, the lowerbound of the bordered [(p^{n}+1)/2(q^{n}1)/2  1(?)] % (p^{n}p^{2})/(m1) ain't 0 so has overlap with [1..x (p^{n}p^{2})/(p1)]. 

Forced imperfection 
Due to "natural overlap" in the numberanges of the border and the bordered two imperfections are forced upon this scheme 

The samples shown below suggest it might be not possible to find the above intended regular borders, so the augmenting hypercube might neccesairily have an off summing line (+/1) the border symmetry suggest at least one other (curently I don't see the "proof" of this, future upload might hold this) 
Case study  

The following case studies generate conditions for the main part of the border however I think it also applies to latin content of the Augmenting hypercube 

The Concentric Bordered square  
The above construction is general and so also applicable for the square This section provide some notes for this situation 

Magic line pair [a,..,b] [a,..,m+1b] 
The set of permutation [a,..,b] [a,..,m+1b] in the range [1..m] sum automatically to the magic sum m(m+1)/2 and are the filling of the first row resp first column the pair (b,m+1b) thus automatically fulfill the subdiagonal restriction, further content of the last row and column follow from the 1agonal and maindiagonal condition 

The Concentric Bordered cube  
The following section is a case study of the situation for the cube to provide understanding of further conditions which might be genarisable to the highter dimensions 

Magic Square triplet {S_{f},S_{l},S_{t}} 
The triplet of squares to the front, left and top share the number at position [0,0,0] further the line [i,0,0] is common to the front and top square, while the line [0,i,0] is shared by the front and left facial squares of the target cube and [0,0,i] is shared by the left and top facial squares of the target cube 

aside from the above, note that the back square is a complemented borderflipped copy of the front square which connects the leftside square line [0,i,m1] with the front squares line [m1,i,0] and the top square line [i,0,m1] with the front line [i,m1,0] further reflection also connect the left squares bottom line [0,m1,i] with the top squares line [m1,0,i] in a simular manner. 

thus summarising in the following conditions: S_{l}[0,i,0] = S_{f}[0,i,0], S_{l}[0,0,i] = S_{t}[0,0,i], S_{t}[i,0,0] = S_{f}[i,0,0]; i = 0..m1 S_{l}[0,i,m1] = m^{2}+1  S_{f}[m1,i,0], S_{t}[i,0,m1] = m^{2}+1  S_{f}[i,m1,0]; i = 2..m2 S_{l}[0,0,m1] = m^{2}+1  S_{f}[m1,m1,0], S_{l}[0,m1,m1] = m^{2}+1  S_{f}[m1,0,0] S_{t}[0,0,m1] = m^{2}+1  S_{f}[m1,m1,0], S_{t}[m1,0,m1] = m^{2}+1  S_{f}[0,m1,0] S_{l}[0,m1,i] = m^{2}+1  S_{t}[m1,0,i]; i = 2..m2 
Concentric Square Border  

01 02 18 21 23 19 ............ 07 20 ............ 06 22 ............ 04 03 24 08 05 25 
div 5 border 0 0 3 4 4 3 ....... 1 3 ....... 1 4 ....... 0 0 4 1 0 4 
mod 5 border 0 1 2 0 2 3 ....... 1 4 ....... 0 1 ....... 3 2 3 2 4 4 
Note in this sample the imperfection of the first and last row (noticable in the div 5 square) simularly the 2 imperfections in the Trumps border in the front and back facial squares 

Trumps Concentric Border: {S(B_{5,3}) = (53)(5^{3}+1)/2 = 126}  
033 022 113 042 105 002 089 106 019 099 085 082 004 119 025 098 090 006 112 009 097 032 086 023 077 
095 080 111 011 018 091 ................. 035 016 ................. 110 005 ................. 121 108 046 015 115 031 
100 017 048 034 116 114 ................. 012 083 ................. 043 008 ................. 118 010 109 078 092 026 
038 102 003 125 047 081 ................. 045 030 ................. 096 087 ................. 039 079 024 123 001 088 
049 094 040 103 029 027 037 020 107 124 101 044 122 007 041 117 036 120 014 028 021 104 013 084 093 
mod 25 Border:  
07 21 12 16 04 01 13 05 18 23 09 06 03 18 24 22 14 05 11 08 21 06 10 22 01 
19 04 10 10 17 15 ............ 09 15 ............ 09 04 ............ 20 07 20 14 14 05 
24 16 22 08 15 13 ............ 11 07 ............ 17 07 ............ 17 09 08 02 16 00 
12 01 02 24 21 05 ............ 19 04 ............ 20 11 ............ 13 03 23 22 00 12 
23 18 14 02 03 01 11 19 06 23 00 18 21 06 15 16 10 19 13 02 20 03 12 08 17 
div 25 Border:  
1 0 4 1 4 0 3 4 0 3 3 3 0 4 0 3 3 0 4 0 3 1 3 0 3 
3 3 4 0 0 3 ....... 1 0 ....... 4 0 ....... 4 4 1 0 4 1 
3 0 1 1 4 4 ....... 0 3 ....... 1 0 ....... 4 0 4 3 3 1 
1 4 0 4 1 3 ....... 1 1 ....... 3 3 ....... 1 3 0 4 0 3 
1 3 1 4 1 1 1 0 4 4 4 1 4 0 1 4 1 4 0 1 0 4 0 3 3 