In fact, any pair of elements you choose will always fulfill the condition (as there will always be one greater than the other one, and both between the smallest and the greatest, including these two!), thus you have to calculate the number of sets with two elements out of a set with $\,n\,$ elements, i.e.
$$\binom{n}{2}=\frac{n!}{2!(n-2)!}=\frac{(n-1)n}{2}$$
ADDED: as Binn points out in the comment below, the condition $\,a_1\leq a_i\leq a_j\leq a_n\,$ implies that we can choose the same element twice, even if the wording "...choose pairs.." can be misleading.
This being the case, we must add $\,n\,$ cases to the above number, each for every element chosen twice, and we thus get the number
$$\frac{(n-1)n}{2}+n=\frac{n(n+1)}{2}$$
Suppose that there are $n$ such sets $A_1,A_2,\ldots,A_n$, represented by indicator vectors $\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n\in\mathbb{F}_2^n$. Equip $\mathbb{F}_2^n$ with the usual inner product $\langle\_,\_\rangle$.
We already know that the vectors $\mathbf{a}_1,\mathbf{a}_2,\ldots,\mathbf{a}_n$ are linearly independent. Therefore, they span $\mathbb{F}_2^n$. Thus, the vector $\boldsymbol{1}:=(1,1,\ldots,1)$ can be written as
$$\mathbf{a}_{j_1}+\mathbf{a}_{j_2}+\ldots+\mathbf{a}_{j_k}$$
for some $j_1,j_2,\ldots,j_k\in\{1,2,\ldots,n\}=:[n]$ with $j_1<j_2<\ldots<j_k$. If $k<n$, then there exists $r\in[n]$ such that $r\neq j_\mu$ for all $\mu=1,2,\ldots,k$. That is,
$$1=\langle \mathbf{a}_r,\boldsymbol{1}\rangle =\sum_{\mu=1}^k\,\langle \mathbf{a}_{j_\mu},\mathbf{a}_r\rangle=0\,,$$
which is a contradiction. Therefore, $k=n$, whence
$$\boldsymbol{1}=\sum_{j=1}^n\,\mathbf{a}_j\,.\tag{*}$$
Consequently, each element of $[n]$ belongs in an odd number of $A_1,A_2,\ldots,A_n$, whence at least one of the sets $A_1,A_2,\ldots,A_n$.
Furthermore, it is not difficult to show that every element of $[n]$ must belong in at least two of the $A_i$'s. (If there exists an element of $[n]$ belonging in exactly one $A_j$, then you can show that there are at most $n-2$ possible $A_i$'s.) Let $d_j$ be the number of sets $A_i$ such that $j\in A_i$. Then
$$\sum_{j=1}^n\,d_j=3n\,.\tag{#}$$
Note that $d_j\geq 2$ for all $j\in[n]$.
From (*), we conclude that $d_j\geq 3$ for every $j\in[n]$. However, (#) implies that $d_j=3$ for all $j\in[n]$; i.e., every element of $[n]$ must be in exactly three of the $A_i$'s. Write $\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n$ for the standard basis vectors of $\mathbb{F}^n_2$. We see that
$$\mathbf{e}_j=\mathbf{a}_p+\mathbf{a}_q+\mathbf{a}_r$$
where $j$ is in $A_p$, $A_q$, and $A_r$. This shows that
$$A_p=\{j,x,y\}\,,\,\,A_q=\{j,y,z\}\,,\text{ and }A_r=\{j,z,x\}$$
for some $x,y,z\in[n]$. Since $x$ already belongs in $A_p$ and $A_r$, it must be belong in another $A_s$. Clearly, $A_s$ must be equal to $\{x,y,z\}$. From here, we conclude that the four elements $j,x,y,z$ belong in exactly four of the $A_i$'s, which are $\{j,x,y\},\{j,y,z\},\{j,z,x\},\{x,y,z\}$. The rest is easy.
Best Answer
I think the answer is: $$f(n)=n+2 \;\;\;\;\;\;\; \mathrm{for} \; n \geq 3.$$
This can be proved by induction.
Since the OP has already established that $f(3)=5$ and $f(4)=6$. The statement is true for $n=3$ and $n=4$.
We need only prove that if the statement is true for $n=k$, then it is true for $n=k+2.$
For simplicity, let us see how to prove that if $f(4)=6$, then $f(6)=8$. The general case is just the same.
Let us fill up the smallest and the largest sets, i.e. $A_1$ and $A_6$ first.
So we may let $$A_1=\{ 1\}$$ and $$A_{6}=\{ 2, 3, \cdots, 7\}.$$ $$$$ Next we consider $A_2$ to $A_5$.
Obviously each $A_i$ must contain at least one element other than $1, 2, \cdots, 7$. We must introduce at least one more element, say $8$.
Therefore at least $8=6+2$ elements are needed.
To see that the addition of $8$ is enough, we may reason as follows:
First assign $8$ to each each $A_i, \; i=2, 3,4, 5$. Thus
$A_2 = \{8, x \}$, $A_3 = \{8, x, x \}$, $A_4 = \{8, x, x, x \}$, $A_5= \{8, x, x, x, x\}$.
where the $x$'s are elements to be filled in.
We may also write it as follows:
$$A_2=\{8 \}\cup B_1, A_3=\{8 \}\cup B_2, A_4=\{8 \}\cup B_3 \; \mathrm{and \; A_5=\{8 \}\cup B_4}$$
where $|B_i|=i$ and $B_i \not\subset B_j$ whenever $i \neq j.$
Thus to fill up $B_2$ to $B_5$ is just the same question asked with $n=4.$
Can we fill up $B_2$ to $B_5$ by using the elements of $\{2, 3, 4, 5, 6, 7\}$ so that no extra element is needed?
The answer is yes. This is because $f(4)=6$ and there are exactly $6$ elements in $\{2, 3, 4, 5, 6, 7\}$.
For concreteness, we may let $$B_1=\{ 2\}, B_2=\{3, 4\}, B_3=\{3, 5, 6 \} \; \mathrm{and} \; B_4=\{4, 5, 6, 7 \}$$ and finally obtain $$A_1=\{1\}, A_2=\{8, 2\}, A_3=\{8, 3, 4\}, A_4=\{8, 3, 5, 6 \}, A_5=\{8, 4, 5, 6, 7 \} \; \mathrm{and} \; A_6=\{2, 3, 4, 5, 6, 7 \}$$
Thus we have proved that $f(6)=8$ if $f(4)=6$.
Same arguments obviously apply for all $n$.
Therefore if the statement is true for $n=k$, then it is true for $n=k+2.$
By the principle of mathematical induction, $$f(n)=n+2 \;\;\;\;\;\;\; \mathrm{for} \; n \geq 3.$$