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Note that if $k_i\geq \lambda_i$ for all $i\in \{1,\dots,l\}$ the term $$ \binom{n}{k_1,\dots ,k_l}^{-1} $$ can only be maximal if for any $i,j\in \{1,\dots,l\}$ with $i\neq j$ we have $k_i=\lambda_i$ or $k_j=\lambda_j$. Hence to get the maximal value we need to put $k_i=\lambda_i$ for all but one $i\in \{1,\dots,l\}$. Hence we have for $n$ large enough $$ \binom{n}{k_1,\dots ,k_l}^{-1}\leq \max_{j\in \{1,\dots,l\}} \frac{(n+\lambda_{j}-\sum_{i=1}^l \lambda_i)!\prod_{i=1}^l \lambda_i!}{\lambda_{j}!n!}\leq \frac{(n+\lambda_{max}-\sum_{i=1}^l \lambda_i)!\prod_{i=1}^l \lambda_i!}{\lambda_{max}!n!} $$ A double counting argument yields $$ \sum_{\substack{k_1+\dots+k_l=n\\k_i\geq \lambda_i}} 1=\binom{n-\sum_{l=1}^l\lambda_i+l-1}{l-1} $$ Hence we have $$ \sum_{\substack{k_1+\dots+k_l=n\\k_i\geq \lambda_i}} \binom{n}{k_1,\dots, k_l}^{-1}\leq \binom{n-\sum_{l=1}^l\lambda_i+l-1}{l-1}\frac{(n+\lambda_{max}-\sum_{i=1}^l \lambda_i)!\prod_{i=1}^l \lambda_i!}{\lambda_{max}!n!}\leq \frac{\prod_{i=1}^l \lambda_i!}{(l-1)!\lambda_{max}! }(n+l-1)^{-\sum_{i=1}^l \lambda_i+\lambda_{max}+(l-1)}. $$ For example for $\lambda_1=\dots=\lambda_l=2$ we get $$ \sum_{\substack{k_1+\dots+k_l=n\\k_i\geq \lambda_i}} \binom{n}{k_1,\dots, k_l}^{-1}\leq \frac{2^{l-1}}{(l-1)!} (n+l-1)^{-(l-1)}. $$