This is not a complete answer (only gives the answer for p=2,3,5) but it is also too long to add as a comment!
Known results concerning similar questions as yours suggest that the nilpotency class of the whole group is a function depending on both the number of generators of the whole group as well as the prime p. For example there is a group of exponent 5 in which every 2-generated subgroup is nilpotent of class at most 6 and every 3-generator subgroup is of nilpotent class at most 8 and for m>3, every m-generated subgroup is of nilpotent class at most 2m. see the following
Newman, M. F. (5-ANUM); Vaughan-Lee, Michael (4-OXCH) Engel-4 groups of exponent 5. II. Orders. Proc. London Math. Soc. (3) 79 (1999), no. 2, 283–317.
Of course you are asking of the nilpotency class of a finite p-group in which every 3-generator (2-generator) subgroup is at most p-1, so the above example does not apparently answer your question.
The case 2-generator has been studied extensively but not for the nilpotency $p-1$ as you mentioned. The case $p=2$ implies the abelian and $p=3$ implies the nilpotency class 3 (as the group will be 2-Engel). The first non-trivial case is $p=5$ in which the group will be 4-Engel. If we assume 3-local assumption (that is every 3-generator subgroup is nilpotent of class $5-1=4$, then answer is positive and the nilpotency class of the whole group is at most 6. It follows from Corollary of
H. Heineken, Bounds for the nilpotency class of a group, J. London Math. Soc. 37 (1962), 456–458.
There is a related question. One may look at the following:
Abdollahi, Alireza, Certain locally nilpotent varieties of groups, Bull. Austral. Math. Soc. 67 (2003), no. 1, 115–119.