Added: The corrected problem is a good deal harder. I’ll denote the set $\{1,\dots,n\}$ by $[n]$. Call a subset of $[n]$ good if it contains at least three consecutive integers, and bad otherwise. Let $g(n)$ be the number of good subsets of $[n]$, and let $b(n)=2^n-g(n)$ be the number of bad subsets of $[n]$. Consider a bad subset $A$ of $[n]$: both $A$ and $A\cup\{n+1\}$ are bad subsets of $[n+1]$ unless $n-1,n\in A$. In that case $n-2\notin A$, so $A\setminus\{n-1,n\}$ is a bad subset of $[n-3]$. Every bad subset of $[n+1]$ is either a bad subset of $[n]$ or a set of the form $A\cup\{n+1\}$ for some bad $A\subseteq[n]$ of the form $B\cup\{n-1,n\}$ for some bad $B\subseteq[n-3]$. There are $b(n)$ bad subsets of $[n]$, and there are $b(n)-b(n-3)$ bad subsets of $[n]$ of the form $B\cup\{n-1,n\}$ for some bad $B\subseteq[n-3]$, so we have the recurrence $$b(n+1)=2b(n)-b(n-3)\;.\tag{1}$$
This implies that
$$\begin{align*}g(n+1)&=2^{n+1}-b(n+1)\\
&=2^{n+1}-\Big(2b(n)-b(n-3)\Big)\\
&=2^{n+1}-2\Big(2^n-g(n)\Big)+2^{n-3}-g(n-3)\\
&=2g(n)-g(n-3)+2^{n-3}\;,\tag{2}
\end{align*}$$
giving us a reasonably nice recurrence for $g$ as well. It can also be written as $$g(n+1)=2g(n)+b(n-3)\;,$$ showing that $g(n)$ more than doubles at each stage.
For the curious, here are the first few values of $g(n)$ and $b(n)$:
$$\begin{array}{r|cc}
n&3&4&5&6&7&8&9\\ \hline
g(n)&1&3&8&20&47&107&238\\
b(n)&7&13&24&44&81&149&274
\end{array}$$
Note that they are not given by the expression $$(n-2)\sum_{k=0}^{n-3} \binom{n-3}k - (n-3)=(n-2)2^{n-3}-(n-3)\;,$$ which gives $10$ for $n=5$. (The eight good subsets of $[5]$ are $\{1,2,3\},\{1,2,3,4\},\{1,2,3,5\}$, $\{1,2,3,4,5\},\{2,3,4\},\{2,3,4,5\}$, and $\{3,4,5\}$.)
The sequence of $g(n)$’s is A050231 in OEIS, which gives my recurrence $(2)$ and a linear recurrence,
$$g(n)=3g(n-1)-g(n-2)-g(n-3)-2g(n-4)\tag{3}$$
with initial conditions $g(0)=g(1)=g(2)=0,g(1)=1$. (These initial conditions also work for $(2)$.) It does not give a closed form.
The sequence of $b(n)$’s is also in OEIS; these are the Tribonacci numbers, OEIS A000073, satisfying
$$b(n)=b(n-1)+b(n-2)+b(n-3)\;,$$
with indices shifted so that the initial conditions are $b(0)=1,b(1)=2,b(2)=4$. They have a known closed form, but it’s not very useful:
$$g(n)=\left\lfloor\frac{\gamma(\alpha+\beta+1)^{n+2}}{3^{n+1}(\gamma^2-2\gamma+4)}+\frac12\right\rfloor\;,$$
where
$$\begin{align*}
\alpha&=\big(19+3\sqrt{33}\big)^{1/3}\;,\\
\beta&=\big(19-3\sqrt{33}\big)^{1/3}\;,\text{ and}\\
\gamma&=\big(586+102\sqrt{33}\big)^{1/3}\;.
\end{align*}$$
It appears to me that you’ll probably have to use one of the recurrences.
Answer to the problem as originally stated:
There are $$1+n+\binom{n}2=1+n+\frac{n(n+1)}2=\frac{(n+1)(n+2)}2$$ subsets containing fewer than $3$ elements, so the number of sets containing at least $3$ elements is $$2^n-\frac{(n+1)(n+2)}2\;,$$ or, more concisely, $\displaystyle2^n-\binom{n+2}2$.
To compute this modulo $10^9+7$, you may be able to make use of the fact that
$$2^{30}=1,073,741,824\equiv 73,741,817\pmod{10^9+7}\;.$$
Better yet, $10^9+7$ is known to be prime, so by Fermat’s little theorem $2^{10^9+6}\equiv 1\pmod{10^9+7}$.