A little more can be proven, namely, if we only assume that $p$ is a bounded sequence (say of complex numbers) verifying the given equations for all $n\ge N$, then $p(k)=\lambda(-1)^k$ for all $k\in\mathbb{N}$.

Let $u$ be the linear form on $\mathbb{C}[z]$ (as linear space) defined by $\langle u, z^k\rangle:=p(k)$ for any $k\in\mathbb{N}
$. Then, since for any $n$
$$\sum _{i=0} ^{n} p(i)\binom {n}{i}=\big\langle u, \sum _{i=0} ^{n} \binom {n}{i} z^i\big\rangle= \langle u, (z+1)^n\rangle$$
verifying the system of the given equations for all $n\ge N$ may be translated as: $u$ vanishes on the ideal $I$ generated by the polynomial $(z+1)^N$, that is, it belongs to the annichilator $I^{\perp}$ of $I$ as linear subspace. Note that by linear algebra, $\dim I^{\perp}=\dim (\mathbb{C}[z]/I)=N$.

Let $u_j$ be linear form that takes the value of the $j$-th derivative at $-1$ , namely $$\langle u_j, P\rangle:=D^jP(-1), \quad\mathrm{for \; } j=0,\dots,N-1,$$ for any $P\in \mathbb{C}[z].$
Clearly, $\{u_0,u_1,\dots,u_{N-1}\}$ are linearly independent element of $I^\perp$, hence a basis. Note that for all $j$ the sequence $p_j(k), (k=0,1,\dots)$
$$p_j(k):=\langle u_j, z^k\rangle:= (-1)^{k-j} k(k-1)\dots(k-j+1) $$
is $(-1)^k$ times a polynomial in $k$ of degree $j$.

As a conclusion, any $p\in \mathbb{C}^\mathbb{N}$ such that for all $n \geq N$

$\sum _{i=0} ^{n} p(i)\binom {n}{i}=0$

must be a linear combination of $p_j$, thus $(-1)^k$ times a polynomial in $k$,

and if $p$ is bounded, it must be a multiple of $p_0$.