$\newcommand\si\sigma$This is not true in general. Indeed, $$B(n_1)=B_\si(n_1):=E(((1-t)N_2+t(N-n_1))^\si|N_1=n_1)$$ for random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$.
If the inequality in question were true for all real $\si>1$, then we would have $$d_t(n_1):=M_\infty(n_1)-M_\infty(n_1+1)\le1 \tag{1}\label{1}$$ for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, where $$M_\infty(n_1):=\lim_{\si\to\infty}B_\si(n_1)^{1/\si}=(1-t)m(n_1)+t(N-n_1), $$ where in turn $m(n_1)$ is the maximum of the support set of the conditional distribution of $N_2$ given $N_1=n_1$. So, $$d_t(n_1)=(1-t)(m(n_1)-m(n_1+1))+t.$$ So, if \eqref{1} were true for all $t\in(0,1)$ and all $n_1\in\{0,\dots,N-1\}$, then we would have $$d_0(n_1)=m(n_1)-m(n_1+1)\le1 \tag{2}\label{2}$$ for all $n_1\in\{0,\dots,N-1\}$. However, it is easy to construct random variables $N_1$ and $N_2$ with values in the set $\{0,\dots,N\}$ such that $N_1+N_2\le N$ and $E(N_2|N_1)=p(N-N_1)$ such that \eqref{2} fails to hold for some $n_1\in\{0,\dots,N-1\}$. $\quad\Box$