Asymptotic bound of a simple alternating binomial sum

For some $p$ and $n$, such largest $N$ does not have to exist.

Indeed, the left-hand side of your inequality equals $$\begin{aligned}L_p(n,N) &:=\sum_{j=0}^n p^{n-j}(-1)^j \binom Nj +\sum_{j=n+1}^N (-1)^j \binom Nj \\ &=\sum_{j=0}^n p^{n-j}(-1)^j \binom Nj +\sum_{j=0}^N (-1)^j \binom Nj-\sum_{j=0}^n (-1)^j \binom Nj \\ &=\sum_{j=0}^n p^{n-j}(-1)^j \binom Nj +(1-1)^N-\sum_{j=0}^n (-1)^j \binom Nj \\ &=\sum_{j=0}^n p^{n-j}(-1)^j \binom Nj -\sum_{j=0}^n (-1)^j \binom Nj \\ &=\sum_{j=0}^n (p^{n-j}-1)(-1)^j \binom Nj . \end{aligned}\tag{1}\label{1}$$

So, if e.g. $p=2$ and $n=5$, then $$L_p(n,N)=L_2(5,N)=\frac{1}{24} \left(N^4-18 N^3+131 N^2-474 N+744\right)>0$$ for all $N$.

More generally, it follows from \eqref{1} that $L_p(n,N)$ is a polynomial in $N$ of degree $n-1$, with the leading coefficient $(-1)^{n-1}\frac{p-1}{(n-1)!}$. So, the set of the values of $N$ for which $L_p(n,N)>0$ is unbounded for each odd $n$.

Remark: Using the identity $\binom Nj=\binom{N-1}j+\binom{N-1}{j-1}$, one gets the recursion $$L_p(n,N)=L_p(n,N-1)-L_p(n-1,N-1),$$ with $L_p(0,N)=0$ and $L_p(n,0)=p^n-1$.

Since @Iosif Pinelis already gave the answer, this is too long for a comment.

Formula $(1)$ could be obtained simplifying for $N>1$ $$\left(\frac{p-1}{p}\right)^N\, p^n-\frac {(-1)^n }{N p}\binom{N}{n+1}\Bigg((n+1)p -N \, _2F_1\left(1,n-N+1;n+2;\frac{1}{p}\right)\Bigg)$$ where

$$ \, _2F_1\left(1,n-N+1;n+2;\frac{1}{p}\right)=\frac{(n+1)!}{(n-N)! }\,\sum_{k=0}^\infty \frac{(k+n-N)!}{(k+n+1)!}\, p^ {-n}$$

As far as I could see it, the problem is related to the parity of $n$. For example

$$L_3(6,N)=-\frac{1}{60} \left(N^5-30 N^4+415 N^3-3450 N^2+17584 N-43680\right)$$ is positive if $N \leq 9$ $$L_3(7,N)=\frac{1}{360}\left(N^6-39 N^5+715 N^4-8205 N^3+63724 N^2-318276 N+786960\right)$$ is always positive.