We distinguish whether $n$ is odd or even.

i) The case of $n$ odd. Since all the $d_i$'s are odd too, it follows from $11d_5+8d_7 = 3n$ that $d_7 \mid 11d_5$ as well as $d_5 \mid 8d_7$, hence $d_5 \mid d_7$. In view of $d_7 > d_5$, the relations $d_5 \mid d_7 \mid 11d_5$ imply that $d_7 = 11d_5$. Substituting this into $11d_5 + 8d_7 = 3n$, we obtain $99d_5 = 3n$ or $33d_5 = n$. Thus the four numbers 1, 3, 11 and 33 are divisors of $n$, more exactly all its divisors smaller than 50, since the fifth divisor $d_5$ satisfies $d_5 = d_3 + 50 > 50$. Consequently, it holds that $d_1 = 1$, $d_2 = 3$, $d_3 = 11$, $d_4 = 33$, $d_5 = d_3 + 50 = 61$, and thus $n = 33d_5 = 33 \cdot 61 = 2013$. The number 2013 is satisfactory indeed, because its first small divisors as indicated in the last sentence and moreover, the subsequent divisors are $d_6 = 61 \cdot 3$ and $d_7 = 61 \cdot 11$; hence $d_7 = 11d_5$ as required.

ii) The case of $n$ even. Now the equality $11d_5 + 8d_7 = 3n$ implies that $2 \mid d_5$ and hence $2 \mid d_5 - 50 = d_3$ as well. Since $d_1 = 1$, $d_2 = 2$ and $d_3 \neq 3$, we conclude that either $d_3 = 4$, or $d_3 = 2t$, with some integer $t > 2$. But the last is impossible (otherwise $t$ is a divisor of $n$ with $d_2 < t < d_3$, a contradiction), Therefore, we have $d_3 = 4$, $d_5 = d_3 + 50 = 54$ and hence 3 is a divisor of $n$ between $d_2$ and $d_3$, a contradiction. In this way, the nonexistence of any satisfactory even $n$ is proven.

Answer. The problem has the only solution $n = 2013$.

The divisors $d_5$ and $d_7$ of $n$, with $d_5 < d_7$, can be represented as $d_5 = n/x$ and $d_7 = n/y$, where $x$ and $y$ ($x > y$) are some positive divisors of $n$ again. Substituting this into $11d_5 + 8d_7 = 3n$, we obtain (after cancelling $n$) an equation $11/x + 8/y = 3$ which can be solved in a standard way, for example by a simple factorization:
$$8x = y(3x - 11) \Leftrightarrow 8(3x - 11) + 88 = 3y(3x - 11) \Leftrightarrow (3x - 11)(3y - 8) = 88.$$
The first equation implies that $3x - 11 > 0$ and hence $3y - 8 > 0$ as well. Note that $x \ge y + 1$ yields $3x - 11 \ge 3y - 8 > 0$. Taking into account the prime factorization $88 = 2^3 \cdot 11$, we conclude that the ordered pair $(3x - 11, 3y - 8)$ of factors must belong to the following set
$$\{(88, 1), (44, 2), (22, 4), (11, 8)\}.$$
However, congruences modulo 3 imply the only two pairs (88, 1) and (22, 4) are admissible. The corresponding pairs $(x, y)$ are (33, 3) and (11, 4), respectively.

If $(x, y) = (33, 3)$, then $d_5 = n/33$ (and $d_7 = n/3$), thus 1, 3, 11 and 33 are divisors of $n$ which leads (as in the above solution) to the solution $n = 2013$.
If $(x, y) = (11, 4)$, then $d_5 = n/11$ and $d_7 = n/4$, thus 1, 2, 4, 11, 22 and 44 are divisors of $n$, which contradicts $d_5 > 50$.