Integrate: $\int_0^1 \mathrm{d}x_1 \int_0^1 \mathrm{d}x_2 \ldots \int_0^1 \mathrm{d}x_n \delta\left( \sum_{i=1}^n k_i x_i \right)$

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \begin{align} &\color{#0000ff}{\large\int_{0}^{1}\dd x_{1}\int_{0}^{1}\dd x_{2}\ldots\int_{0}^{1}\dd x_{n}\,\delta\pars{\sum_{i = 1}^{n}k_{i}x_{i}}} \\[3mm]&=\int_{0}^{1}\dd x_{1}\int_{0}^{1}\dd x_{2}\ldots\int_{0}^{1}\dd x_{n} \int_{-\infty}^{\infty}\expo{\ic q\sum_{i = 1}^{n} k_{i}x_{i}}\,{\dd q \over 2\pi} \\[3mm]&= \int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n}\int_{0}^{1} \expo{\ic qk_{i}x_{i}}\,\dd x_{i} = \int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n} {\expo{\ic qk_{i}} - 1 \over \ic qk_{i}} = \int_{-\infty}^{\infty}{\dd q \over 2\pi}\prod_{i = 1}^{n} \expo{\ic qk_{i}/2}\,{2\ic\sin\pars{qk_{i}/2} \over \ic qk_{i}} \\[3mm]&=\color{#0000ff}{\large% {2^{n - 1} \over \pi}\int_{-\infty}^{\infty}{\dd q \over q^{n}}\prod_{i = 1}^{n} {\expo{\ic qk_{i}/2}\sin\pars{qk_{i}/2} \over k_{i}}} \end{align} I don't see any further reduction unless we know something else about the $\braces{k_{i}}$.

$\delta(x)$ is not really a function in classical sense. For the purpose of deriving an expression without involving the concept of distribution, we will treat it as some sort of derivative of a step function. Assume all $k_i \ne 0$, let

$$\lambda_i = |k_i|,\quad y_i = \begin{cases}x_i,& k_i > 0\\1-x_i,& k_i < 0\end{cases}, \quad K = \left|\prod_{i=1}^n k_i \right| \quad\text{ and }\quad L = \sum_{k_i < 0} |k_i| $$

We have $$\delta\left( \sum_{i=1}^n k_i x_i \right) = \delta\left(\sum_{i=1}^n\lambda_i y_i - L\right) = \frac{d}{dL} \theta\left(L - \sum_{i=1}^n\lambda_i y_i\right)$$ where $\quad \displaystyle \theta(x) = \begin{cases} 1, &x > 0\\0, & x \le 0\end{cases}\quad$ is the step function. We can evaluate the integral as

$$\begin{align}\mathcal{I} =& \frac{d}{dL} \left[ \int_0^1 dy_1 \cdots \int_0^1 dy_n \theta\left( L - \sum_{i=1}^n \lambda_i y_i\right) \right]\\ =& \frac{d}{dL} \left[ \left( \int_0^\infty - \int_1^\infty \right) dy_1 \cdots \left( \int_0^\infty - \int_1^\infty \right) dy_n \theta\left( L - \sum_{i=1}^n \lambda_i y_i\right) \right]\\ =& \frac{d}{dL} \left[ \int_0^\infty dy_1 \cdots \int_0^\infty dy_n \sum_{0\le \epsilon_1, \ldots, \epsilon_n \le 1 } (-1)^{\sum_{i=1}^n \epsilon_i} \theta\left( \left(L - \sum_{i=1}^n\lambda_i\epsilon_i\right) - \sum_{i=1}^n \lambda_i y_i\right) \right] \end{align}$$

Notice the integral

$$\int_0^\infty dy_1\cdots\int_0^\infty dy_n \theta\left( X - \sum_{i=1}^n \lambda_i y_i\right)$$ is the volume of a simplex and equal to $\begin{cases}\frac{X^n}{n!K},& X > 0\\ \\0, &\text{otherwise}\end{cases}$, we have

$$\begin{align}\mathcal{I} =& \frac{1}{n!K} \frac{d}{dL} \left[ \sum_{0\le \epsilon_1, \ldots, \epsilon_n \le 1 } (-1)^{\sum_{i=1}^n \epsilon_i} \left(L - \sum_{i=1}^n\lambda_i\epsilon_i\right)^n \right]\\ =& \frac{1}{(n-1)!K} \sum_{0\le \epsilon_1, \ldots, \epsilon_n \le 1 } (-1)^{\sum_{i=1}^n\epsilon_i} \left(L - \sum_{i=1}^n\lambda_i\epsilon_i\right)^{n-1}\\ =& \frac{1}{(n-1)!K} \sum_{\stackrel{0\le \epsilon_1,\ldots,\epsilon_n \le 1;}{L - \sum_{i=1}^n \lambda_i\epsilon_i > 0} } (-1)^{\sum_{i=1}^n\epsilon_i} \left(L - \sum_{i=1}^n\lambda_i\epsilon_i\right)^{n-1} \end{align}$$ A the end, $\mathcal{I}$ is a sum of polynomials of the form $\left(L - \text{ ???}\right)^{n-1}$ and the sum only runs over those $???$ which is smaller than $L$.

The integral looks like it's related to box splines. The idea of box splines is to express b-spline basis functions as areas of planar slices of hypercubes. John's comment has a good explanation of the quadratic case. As he said, you get a piecewise quadratic function which is (I think) a b-spline basis function. Look up box splines. You'll find papers by Carl deBoor, among others. The Wikipedia page here looks like a good introduction, and it has a long list of references.