Probability for rolling $n$ dice to add up to at least (a specific sum)

Consider that $$ \eqalign{ & {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 1} \le {\rm integer}\;y_{\,j} \le m \hfill \cr y_{\,1} + y_{\,2} + \; \cdots \; + y_{\,n} = s \hfill \cr} \right.\quad \Rightarrow \cr & \Rightarrow \quad {\rm No}{\rm .}\,{\rm of}\,{\rm solutions}\,{\rm to}\;\left\{ \matrix{ {\rm 0} \le {\rm integer}\;x_{\,j} \le m - 1 \hfill \cr x_{\,1} + x_{\,2} + \; \cdots \; + x_{\,n} = s - n \hfill \cr} \right. = \cr & = N_{\,b} (s - n,m - 1,n) \cr} $$ where $N_b$ is given by $$ \eqalign{ & N(sum = s,m,n) = N_b (s - n,m - 1,n)\quad = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - n} \over m}\, \le \,n} \right)} { \left( { - 1} \right)^k \binom{n}{k}\binom{s-1-k\,m}{s-n-k\,m} } \cr} $$ as widely explained in this related post.

Note that the second binomial looks equivalent to that in the formula you cited.
But written in this way it has the advantage to implicitly contain the bounds of the sum, which then can be omitted (indicated in brackets).
This has the advantage of simplifying the the algebraic manipulations, and in fact for the cumulative Number we easily get $$ \eqalign{ & N(sum \le S,m,n) = \cr & = \sum\limits_{0\, \le \,s\, \le \,S} {\sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{{s - n} \over m}\, \le \,n} \right)} { \left( { - 1} \right)^k \binom{n}{k}\binom{s-1-k\,m}{s-n-k\,m} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( {\, \le \,n} \right)} {\sum\limits_{\left({0\, \le}\right) \,s\, \left({\,\le \,S}\right)} { \left( { - 1} \right)^k \binom{n}{k} \binom{S-s}{S-s} \binom{s-1-k\,m}{s-n-k\,m} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( {\, \le \,n} \right)} { \left( { - 1} \right)^k \binom{n}{k} \binom{S-k\,m}{S-n-k\,m}} \cr} $$

by using the "double convolution" formula for binomials.

It is a little easier to answer the probability of rolling less than $s$. This is $m^{-n}$ times the number of solutions to $$ d_1+d_2+\dots+d_n+e=s,\\ 1\le d_i\le m,\\ 1\le e $$ which is the coefficient of $x^s$ in $$ m^{-n}(x+x^2+\dots+x^m)^n\cdot (x+x^2+x^3+\dots)=m^{-n}x^{n+1}(1-x^m)^n(1-x)^{-n-1}. $$ Therefore, $$ \boxed{P(\text{roll less than s})=m^{-n}\sum_{i=0}^{\left\lfloor\frac{s-1-n}{m}\right\rfloor}(-1)^i\binom{n}{i}\binom{s-1-im}{n}} $$ For example, with two six sided dice, you get $$ \frac1{36}\Bigg[\binom{s-1}{2}-2\binom{s-7}{2}+\binom{s-13}{2}\Bigg] $$ which checks out.