A generating function solution.
For every $S\subset\{1,2,\ldots,2015\}$ we will write $\Sigma S=\sum_{k\in S}k$.
Let
$$
f(a,x) = \prod_{k=1}^{2015} (1+a^kx) =
\sum_{S\subset\{1,\ldots,2015\}} a^{\Sigma S} x^{|S|}.
$$
Take the average this function over putting $5$th complex roots of unity for $a$. Let $\omega=e^{2\pi i/5}$; then
$$
\frac15\sum_{j=0}^4 f(\omega^j,x) =
\sum_{S\subset\{1,\ldots,2015\}} \left(\frac15\sum_{j=0}^4 \big(\omega^{\Sigma S}\big)^j\right) x^{|S|} =
\sum_{\substack{S\subset\{1,\ldots,2015\}\\\Sigma S\equiv0\pmod5}} x^{|S|}.
\tag{$*$}
$$
On the RHS of $(*)$, the coefficient of $x^n$ is the number of $n$-element sets $S\subset\{1,\ldots,2015\}$ with $\Sigma S\equiv0\pmod5$.
On the other hand,
$$
f(\omega^j,x)= \begin{cases} (1+x)^{2015} & \text{if } j=0 \\
(1+x^5)^{403} & \text{if } j=1,2,3,4 \end{cases}
$$
so on the LHS of $(*)$ the coefficient of $x^n$ is: $\frac15\binom{2015}n$ if $n$ is co-prime with $5$, and
$\frac15\binom{2015}n+\frac45\binom{403}{n/5}$ if $5|n$.
The comments have already very quickly pointed this out, but perhaps I can give an explanation on why they work.
$1$) André Nicolas' comment has already answered that there are $\binom{14}{2}$ subsets of size $2$. This notation is the binomial coefficient. That is, \begin{align}\binom{14}{2} &= \frac{14!}{(14-2)!2!}\\ &= \frac{14 \times 13 \times \color{red}{12} \times \color{red}{11} \times \color{red}{10} \times \color{red}9 \times \color{red}8 \times \color{red}7 \times \color{red}6 \times \color{red}5 \times \color{red}4 \times \color{red}3 \times \color{red}2 \times \color{red}1}{(\color{red}{12} \times \color{red}{11} \times \color{red}{10} \times \color{red}9 \times \color{red}8 \times \color{red}7 \times \color{red}6 \times \color{red}5 \times \color{red}4 \times \color{red}3 \times \color{red}2 \times \color{red}1)(2 \times 1)}\\ &=\frac{14 \times 13}{2}\\ &= 91. \end{align} This is correct because there are $14$ elements in your set, and we wish to choose $2$ of them each time.
$2$) Ethan Hunt's comment has already answered that there are $2^n$ subsets, including the empty set and the entire set itself. To find all subsets of a set, we take the power set of the set. It's easy to prove that the cardinality of the power set is $2^n$, where $n$ is the number of elements of the set. So in your case, there are $2^{14} = 16384$ subsets.
Best Answer
I compute the number as:
I found this as follows:
Let $S_j(n)$ be the number of subsets of $\{1,2,\ldots,n\}$ whose sum is congruent to $j$ modulo $5$.
When $n \geq 5$, since any subset of $\{1,2,\ldots,n\}$ is the union of a unique pair of subsets, one subset of $\{1,2,\ldots,n-5\}$ and one subset of $\{n-4,n-3,n-2,n-1,n\} \equiv \{0,1,2,3,4\} \pmod 5$, we have the recurrence $$S_j(n) = \sum_{i=0}^4 S_i(n-5)S_k(5)$$ where $k$ is the solution to the congruence $i+k \equiv j \pmod 5$.
I implemented this in GAP using the following code (note: GAP uses indices starting at 1):
The above initialises
Swith the boundary conditions, i.e., when $n=5$.The matrix
Mgives the solutions to $i+k \equiv j \pmod 5$.The above just implements the recurrence.
As some quick sanity checks: (a) the code below checks that the counts sum to $2^n$ for any given $n$ (i.e. the number of subsets of $\{1,2,\ldots,n\}$).
(b) we check that the answer is divisible by $2^{400}$ (as pointed out by Ross Millikan).