At least for parameters $\lambda_i$ that are all different, the value of the integral is
$\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\qquad\quad\displaystyle \sum_i\mathrm{e}^{\lambda_i}\prod_{j\ne i}\frac1{\lambda_i-\lambda_j}. $
To prove this formula, one can denote by $J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})$ the integral of interest when there are $n+1$ parameters, hence $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\int_{[0,1]^n}\mathrm{e}^{\lambda_1x_1+\cdots+\lambda_nx_n+\lambda_{n+1}(1-x_1-\cdots-x_n)}\mathbf{1}_{0\le x_1+\cdots+x_n\le1}\text{d}x_1\cdots\text{d}x_n. $ Equivalently, $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\mathrm{e}^{\lambda_{n+1}}K_n(\mu_1,\ldots,\mu_n), $ with $\mu_i=\lambda_i-\lambda_{n+1}$ for every $i\le n$ and $ K_n(\mu_1,\ldots,\mu_n)=\int_{[0,1]^n}\mathrm{e}^{\mu_1x_1+\cdots+\mu_nx_n}\mathbf{1}_{0\le x_1+\cdots+x_n\le1}\text{d}x_1\cdots\text{d}x_n. $ Now, perform the integral along the last coordinate $x_{n}$. The domain of integration is $0\le x_{n}\le 1-x_{1}-\cdots-x_{n-1}$ and $ \int_0^{1-s}\mathrm{e}^{\mu_{n}x_{n}}\mathrm{d}x_{n}=\frac1{\mu_{n}}(\mathrm{e}^{\mu_{n}(1-s)}-1), $ hence, using the shorthand \mu'_i=\mu_i-\mu_n=\lambda_i-\lambda_{n} for every $i\le n-1$, K_{n}(\mu_1,\ldots,\mu_{n})=\frac1{\mu_n}(\mathrm{e}^{\mu_n}K_{n-1}(\mu'_1,\ldots,\mu'_{n-1})-K_{n-1}(\mu_1,\ldots,\mu_{n-1})). This translates back in terms of $J_{n+1}$ and $J_n$ as $ J_{n+1}(\lambda_1,\ldots,\lambda_{n+1})=\frac1{\mu_n}(J_{n}(\lambda_1,\ldots,\lambda_{n})-J_{n}(\lambda_1,\ldots,\lambda_{n-1},\lambda_{n+1})), $ Starting from $ J_2(\lambda_1,\lambda_2)=\mathrm{e}^{\lambda_1}\frac1{\lambda_1-\lambda_2}+\mathrm{e}^{\lambda_2}\frac1{\lambda_2-\lambda_1}, $ this yield the desired formula through a recursion over $n$.