Let me try to write an answer. As I said in comments, Higman's conjecture seems still to be open, and does not, in any case, predict a precise formula for $k(U)$.

As for the other question, as clarified in comments, the question is intended to ask whether $k(U) = [G:B]$, where $B = N_{G}(U).$ I point out that when $n > 1$, it is never the case that $k(U) = [G:B].$

To do this, I note that $[G:B] \geq 1+|U|$, while it is clearly the case that $k(U) \leq |U|.$

The Sylow $p$-subgroup $U$ of $G$ has a conjugate $U^{x}$ with $U \cap U^{x} = 1$, (just take $U^{x}$ to be the subgroup of $G$ consisting of all lower unitriangular matrices). Then $B \cap B^{x}$ is a $p^{\prime}$-groups since $B$ has onlyy one Sylow $p$-subgroup. Hence $|BxB| = |BxBx^{-1}| = |B||BxBx^{-1}|/|B \cap xBx^{-1}| \geq |U||B|$, so that $[G:B] \geq |U|$. But $[G:B]$ is coprime to $p$, so that $[G:B] \geq 1+|U| > |U| \geq k(U)$, as claimed.

Let me try to write an answer. As I said in comments, Higman's conjecture seems still to be open, and does not, in any case, predict a precise formula for $k(U)$.

As for the other question, as clarified in comments, the question is intended to ask whether $k(U) = [G:B]$, where $B = N_{G}(U).$ I point out that when $n > 1$, it is never the case that $k(U) = [G:B].$

To do this, I note that $[G:B] \geq 1+|U|$, while it is clearly the case that $k(U) \leq |U|.$

The Sylow $p$-subgroup $U$ of $G$ has a conjugate $U^{x}$ with $U \cap U^{x} = 1$, (just take $U^{x}$ to be the subgroup of $G$ consisting of all lower unitriangular matrices). Then $B \cap B^{x}$ is a $p^{\prime}$-group since $B$ has only one Sylow $p$-subgroup. Hence $|BxB| = |BxBx^{-1}| = |B||xBx^{-1}|/|B \cap xBx^{-1}| \geq |U||B|$, so that $[G:B] \geq |U|$. But $[G:B]$ is coprime to $p$, so that $[G:B] \geq 1+|U| > |U| \geq k(U)$, as claimed.

Later edit: Quicker to say $[G:B] = \prod_{i=2}^{n} \frac{p^{n}-1}{p-1} \geq \prod_{i=1}^{n-1} (p^{i}+1) > \prod_{i=1}^{n-1} p^{i} = |U|$.