If $X_1,\ldots,X_n$ are inedependent uniform$(0,1)$ random variables, then $ {\rm P}(X_1+ \cdots +X_n \leq a) = {\rm volume}(A), $ where $ A = \{ (x_1 , \ldots ,x_n ) \in (0,1)^n :x_1 + \cdots + x_n < a\} . $ For the probability density function of the sum $X_1+ \cdots +X_n$, see this answer.
EDIT: When $0 < a \leq 1$, it holds $ {\rm volume}(A) = \frac{{a^n }}{{n!}}. $
EDIT 2: Probabilistic proof for the case $0 < a \leq 1$. Let $X_1,X_2,\ldots$ be independent uniform$(0,1)$ variables. We want to show that, for any $0 < a \leq 1$, $ {\rm P}(X_1+ \cdots +X_n \leq a) = \frac{{a^n }}{{n!}}. $ This can be easily done by induction, as follows. The case $n=1$ is trivial: ${\rm P}(X_1 \leq a) = a$. Assume that the result is true for $n$, and let $m = n+1$. By the law of total probability, $ {\rm P}(X_1+ \cdots + X_m \leq a) = \int_0^1 {{\rm P}(X_1 + \cdots + X_m \le a|X_m = u)\,du} $ $ = \int_0^a {{\rm P}(X_1 + \cdots + X_m \le a|X_m = u)\,du} = \int_0^a {{\rm P}(X_1 + \cdots + X_n \le a - u)\,du}. $ Hence, by the induction hypothesis, $ {\rm P}(X_1+ \cdots + X_m \leq a) = \int_0^a {\frac{{(a - u)^n }}{{n!}}\,du} = - \frac{{(a - u)^{n + 1} }}{{(n + 1)!}}\bigg|_0^a = \frac{{a^{n + 1} }}{{(n + 1)!}}. $ The result is thus proved.