The answer is $5$. The picture below shows five $1 \times 2$ rectangles mutually intersecting at rectangles with unit area.

It is enough to show that there are no six $1 \times 2$ horizontal or vertical rectangles with such a property. Assume that there are $6$ such rectangles.
Note that it is obvious that the intersection of two rectangles with parallel sides is again a rectangle. In solution first we show that there are at most three vertical and three horizontal rectangles. At the end we show that there are not three horizontal and three vertical rectangles satisfying the intersection property. For these we prove some lemmas.

Lemma 1. Composition of a reflection with respect to a point and a translation in the plane, is a reflection, too.
Proof. Let $O$ be the center of the reflection, and $\vec{v}$ be the vector of the translation. Then by Thales' Theorem it is easy to see the composition is a reflection with respect to point $O'$ such that $OO' = \frac{1}{2}\vec{v}$.

Lemma 2. Let $K$ be a strictly convex shape in the plane. For any real number $r > 0$ and any direction in the plane, there are at most two slices of $K$ with length $r$ parallel to the given direction. (We mean by a slice of $K$, intersection of a line in the plane with $K$. Obviously each slice is a line segment.)
Proof. Assume to the contrary there are three slices of length $r$ parallel to the given direction, say $AA'$, $BB'$ and $CC'$. Suppose that $\overrightarrow{AA'} = \overrightarrow{BB'} = \overrightarrow{CC'}$ and $BB'$ lies between two others.
If $B$ or $B'$ is outside the parallelogram $AA'C'C$, then the other point lies in the interior of $K$ and therefore the intersection of line containing $BB'$ has length greater than $r$.

If $B$ lies on $AC$ and $B'$ lies on $A'C'$, then since $K$ is strictly convex there must be a point of $K$ between lines $AA'$ and $CC'$ outside the parallelogram, contradicts with the length $r$ of slice containing $BB'$.

Lemma 3. There are no four horizontal or four vertical rectangles satisfying the conditions.
Proof. Assume that there are at least $4$ horizontal rectangles. By a horizontal scaling with factor $\frac{1}{2}$ we go to the case that there are $4$ unit squares with sides parallel to the axes, such that mutually intersecting at rectangles of area $\frac{1}{2}$.
Every unit square determined uniquely by its center. It is easy to see that two unit squares with centers $(x, y)$ and $(x', y')$ intersect at a rectangle of area $\frac{1}{2}$ iff
$$(1 - |x - x'|)(1 - |y - y'|) = \frac{1}{2} \quad (*)$$
Locus of points $(x', y')$ satisfying (*) (for fixed $(x, y)$) is boundary of a strictly convex shape in the plane (Look at the shape below). We call such shape an oval with center $(x, y)$.

Suppose that $A = (x_A, y_A)$, $B = (x_B, y_B)$, $C = (x_C, y_C)$ and $D = (x_D, y_D)$ are the centers of these $4$ squares such that $y_A \le y_B \le y_C \le y_D$. Note that oval of $B$ is a translation of oval of $A$ with vector $\vec{AB}$. By lemma $2$ these two ovals intersect in at most two points (Note that these ovals are strictly convex because they are intersection of $4$ strictly convex shapes). These two points must be $C$ and $D$. Since $A$ is center of symmetry of oval of $A$, by lemma $1$, these two intersection points are symmetric with respect to the midpoint of $AB$. Note that $y_B > y_A$, because if $y_A = y_B$, $y_C$ or $y_D$ become less than $y_A = y_B$ contradicts with the minimality of $y_A$. Now we have
$$y_C \le \frac{y_C + y_D}{2} = \frac{y_A + y_B}{2} < y_B$$
Contradicts with $y_B \le y_C$.

Lemma 4. Every horizontal rectangle intersects every vertical rectangle in a $1 \times 1$ square.
Proof. Intersection of such rectangles is a rectangle with sides at most $1$. As intersection of every two rectangles has a unit area, both sides must be unit. □

Using lemma $3$, we get that $3$ rectangles are vertical (V) and other $3$ are horizontal (H). Consider three horizontal rectangles with their centers sorted by $x$. Fix the middle one and call it $R_m$. Call left and right rectangles $R_l$ and $R_r$ respectively. Without loss of generality, we can assume that $R_r$ is not upper than $R_l$. As in the figure below $R_m$, $R_r$ intersect in a rectangle with length $\beta$ and $R_m$, $R_l$ intersect in a rectangle with length $\alpha$. Left side of $R_r$, right side of $R_l$, top side of $R_l$ and bottom side of $R_m$ construct a rectangle named $R_H$. Width of $R_H$ is $w_H$ and height of it is $h_H$. These variables are based on rectangles in (H), we can define the same variables in (V) named $R_V$, $w_V$ and $h_V$.

Referring to the lemma $4$, every two rectangles, one in (V) and one in (H) intersect in a unit square. So every rectangle in (V) must have a horizontal side in $R_H$ this means $h_V \le w_H$. Similarly we can show that $h_H \le w_V$. Without loss of generality, assume that $w_V \le w_H$. This yields:
$$h_H \le w_H$$
Now we can calculate all variables in terms of $\alpha, \beta$ (note that $1 \le \alpha \le \beta \le 2$). Intersection of $R_l, R_r$ is a rectangle with sides $\alpha + \beta - 2, 1 - \frac{1}{\beta} + \frac{1}{\alpha}$ so:
$$(\alpha + \beta - 2) \times \left(1 - \frac{1}{\beta} + \frac{1}{\alpha}\right) = 1 \quad (1)$$
We had $h_H \le w_H$. $h_H = 2 - \frac{1}{\alpha}$ and $w_H = \alpha + \beta - 2$ so:
$$2 - \frac{1}{\alpha} \le \alpha + \beta - 2 \quad (2)$$
Now we change our variables to have simpler results. Assume that:
$$\alpha = a + 1 \quad (0 \le a \le 1)$$
$$\beta = b + 1 \quad (0 \le b \le 1)$$
From (1) we have:
$$\begin{align*} & (b+a)\left(1 - \frac{1}{b+1} + \frac{1}{a+1}\right) = 1 \\ \Rightarrow \quad & b - \frac{b}{b+1} + \frac{b}{a+1} + a - \frac{a}{b+1} + \frac{a}{a+1} = 1 \\ \Rightarrow \quad & b^2 + b - b + \frac{b(b+1)}{a+1} + ab + a - a + \frac{ba}{a+1} + \frac{a}{a+1} = b + 1 \\ \Rightarrow \quad & Q(b) = b^2\left(1 + \frac{1}{1+a}\right) + ba - \frac{1}{a+1} = 0 \end{align*}$$
From (2) we conclude that the above quadratic polynomial must have a root in the interval $[0, 1]$ satisfying the following condition:
$$2 - \frac{1}{a+1} \le a + b \Leftrightarrow \underbrace{1 - \frac{a^2}{a+1}}_{k} = 2 - a - \frac{1}{a+1} \le b$$
$a \in [0, 1]$ so $k \in [0, 1]$. Coefficients of $Q$ are positive so $Q$ is ascending so to prove that $b$ cannot be a root of $Q$, it's enough to show that $Q(k) > 0$. $Q(k) \ge k^2 + ka - \frac{1}{a+1}$ so we prove that $k^2 + ka - \frac{1}{a+1} \ge 0$.
$$\begin{align*} & k^2 + ka - \frac{1}{a+1} \ge 0 \\ \Leftrightarrow \quad & (1 - \frac{a^2}{a+1})^2 + (1 - \frac{a^2}{a+1})a - \frac{1}{a+1} \ge 0 \\ \Leftrightarrow \quad & (a + 1 - a^2)^2 + a(a + 1)(a + 1 - a^2) \ge a + 1 \\ \Leftrightarrow \quad & (a + 1 - a^2)(2a + 1) \ge a + 1 \end{align*}$$
Equivalently,
$$2a(a+1) - a^2(2a+1) = -2a^3 + a^2 + 2a = -a(2a^2 - a - 2) \geq 0$$
$a \in [0, 1]$ so the last inequality is obvious because $2a^2 - a - 2 = (a^2 - a) + (a^2 - 1) - 1$. Thus we showed that there is no $a$ with those algebraic properties. This means there is no $\alpha$ with that geometric properties and this is a big contradiction because $\alpha$ was a length in our figure. Now we can say there are no six $1 \times 2$ rectangles mutually intersecting at rectangles with unit area. This yields there are no $n > 5$ such rectangles.