Using my Maple package BinomSums it is easy to compute the generating function of the left-hand side of your identity (let's call it $u_{n,m,p,n',m',p'}$):
$$\sum_{n,m,p,n',m',p' \geq 0} u_{n,m,p,n',m',p'} x_1^n y_1^m z_1^p x_2^{n'} y_2^{m'} z_2^{p'} = -\frac{(x_2-1)(x_1-1)}{(y_2z_2+x_2-1)(y_1x_2-y_2x_2-y_1-x_2+1)(y_1x_1-y_2x_1+y_2+x_1-1)(y_1z_1+x_1-1)}$$
Thus a rational generating function usually gives all we need about a sequence.
Here is the code:
S := Sum(Sum(Binomial(n1+p1,i1)*Binomial(n2+p2,i2)*Multinomial([i1-p1,m2-i2])*Multinomial([i2-p2,m1-i1]), i1=0..infinity), i2=0..infinity);
vars := [n1,m1,p1,n2,m2,p2]:
gfunS := foldl(Sum, S*mul((t||v)^v, v in vars), seq(v=0..infinity, v in vars));
BinomSums[sumtores](gfunS, u);
Yes, if $\alpha<p$ (if $\alpha>p$, the sum is almost 1). To see this, write
$$
\sum_{\ell = 0}^{\alpha n} \binom{n}{\ell}p^\ell(1-p)^{n-\ell}\leqslant t^{-\alpha n}(pt+(1-p))^n
$$
for every $t\in (0,1]$. Choose a positive $t=t_0$ for which RHS is minimal possible, taking the logarithmic derivative equal to 0 we get $-\alpha/t_0+p/(pt_0+1-p)=0$, $-\alpha p t_0-\alpha(1-p)+pt_0=0$, $pt_0(1-\alpha)=\alpha(1-p)$, $t_0=\frac{\alpha(1-p)}{p(1-\alpha)}$. We see that if $\alpha\leqslant p$, this $t_0$ is indeed in $(0,1]$, thus we get the upper bound $\theta^n$, where
$$
\theta=\frac{p^\alpha(1-p)^{1-\alpha}}{\alpha^\alpha (1-\alpha)^{1-\alpha}}.
$$
Best Answer
Expanding on the previous answers. I'm taking $\lambda$ and $\alpha$ to be constants which do not vary as $n\to\infty$.
If $α<λ/(λ+1)$ then the sum is within a constant of the last term. In fact the largest terms are approximately in geometric progression so you can get it quite accurately by computing the ratio.
If $α>λ/(λ+1)$, almost all of the complete binomial expansion is present, so the sum equals $(1+o(1))(1+λ)^n$.
If $α≈p$, then Russell's normal approximation will be good. (This needs some work to clarify whether the geometric approximation of the lower tail is good right up to the point where the normal approximation begins to be good. I think it is.)