The case of 4 variables

I just derived an closed form for this, here

If we denote the generalised version by $S$ that is,

$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}$

$\displaystyle \begin{aligned} S &= \sum_{k=n}^\infty\frac{1}{k!}\;\left( \sum_{a_1+a_2+\cdots+a_n=k}a_1 a_2\cdots a_n\right) \end{aligned}$

This was achieved by setting $\sum_{i=1}^n a_i =k$ , and what remains to calculate is the inner sum enclosed by brackets.

We start by investigating the lower cases , suppose we have only two variables $a_1,a_2$ with $a_1+a_2=k$ then

$\displaystyle \sum_{a_1+a_2=k}(a_1 a_2) =\sum_{N=1}^{k-1} N(k-N)=\frac{k(k-1)(k+1)}{3!}=\binom{k+1}{3}$

Now if we take the case of $3$ variables where $a_1+a_2+a_3=k$ , we can achieve the sum as :

$\displaystyle \sum_{a_1+a_2+a_3=k} a_1 a_2 a_3 = \sum_{N=1}^{k-2} N\binom{k+1-N}{3}= \frac{k(k-1)(k+1)(k-2)(k+2)}{5!}$

Similarly for $4$ variables it turns out to be ,

$\displaystyle \sum_{a_1+a_2+a_3+a_4=k}a_1 a_2 a_3 a_4 = \frac{k(k-1)(k+1)(k-2)(k+2)(k-3)(k+3)}{7!}$

I believe for the same reason that that ,

$\displaystyle \sum_{a_1+a_2+\cdots+a_n=k}a_1 a_2\cdots a_n = \frac{k}{(2n-1)!}\prod_{m=1}^{n-1}(k-m)(k+m)$

This is indeed tough to prove by induction , but I guess it can be proved due to the great symmetry and pattern this sequence follows. I haven't tried but will try to update a proof on this asap, but till then it's reasonable to conjecture this.

Lastly we have that ,

$\displaystyle \begin{aligned} S &= \sum_{k=n}^\infty \frac{1}{k!} \left(\frac{k}{(2n-1)!}\prod_{m=1}^{n-1}(k-m)(k+m)\right) \\ &= \frac{1}{(2n-1)!}\sum_{k=n}^\infty \frac{1}{k.k!} (k)_n (k)^n \\ &= \frac{1}{(2n-1)!}\sum_{k=n}^\infty \frac{1}{k.k!} \left(\sum_{r=1}^{n}s(n,r)k^r\right) \left(\sum_{t=1}^n {n\brack t}k^t\right) \\ &= \frac{1}{(2n-1)!}\sum_{r,t=1}^n (-1)^{n+r} {n\brack r}{n\brack t}\left(\sum_{k=n}^\infty \frac{k^{r+t-1}}{k!}\right) \end{aligned}$

Now using Dobinski's Formula we have finally,

$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}\\ = \frac{1}{(2n-1)!}\sum_{r=1}^n\sum_{t=1}^n (-1)^{n+r} {n\brack r}{n\brack t} \left[eB_{r+t-1}-\sum_{m=1}^{n-1}\frac{m^{r+t-1}}{m!}\right]$

where $B_n$ is the n-th Bell Number.

Firstly, if we separate the answer into two parts and take the constant term which doesn't have $e$ , we get the constant part as

$\displaystyle \sum_{r=1}^n \sum_{t=1}^n \sum_{m=1}^{n-1} (-1)^{n-r}{n\brack r}{n\brack t}\frac{m^{r+t-1}}{m!}$

A little modification and interchange of sums will give the result in terms of the Pochammer symbol.

$\displaystyle \sum_{m=1}^{n-1} \frac{(m)_n m^{t-1}}{m!} =0$

This sum is eventually equal to zero and is easy to prove by induction.

Thus the answer is :

$\displaystyle \sum_{a_1,a_2,\cdots,a_n=1}^\infty \frac{a_1a_2\cdots a_n}{(a_1+a_2+\cdots+a_n)!}\\ = e\left[\frac{1}{(2n-1)!}\sum_{r=1}^n\sum_{t=1}^n (-1)^{n+r} {n\brack r}{n\brack t} B_{r+t-1}\right]$

For $n=4$ it turns out to be $\displaystyle \boxed{\frac{179}{2520}e}$ making the answer $\boxed{2699}$