Why are so many pins of 8086 active-low? Does it have to do something with the fact that more circuits are able to sink current than to source current, or is there another reason for it?
In the TTL era, active low was the standard.
The old NMOS parts like the 8086 were designed in that time period- note that the input levels are TTL-compatible Vih=2.0V and Vil=0.8V.
Active-low allows wired-OR with open collector/drain outputs (you may have noticed that calling it 'wired-OR' implies negative logic).
Also, the NMOS parts were almost like RTL except the pull-up resistors (loads) were depletion mode n-channel MOSFETs so the negative edges would tend to be sharper with capacitive load. There are no p-channel parts in this technology (which made it simpler and thus higher yield)
CMOS parts (originally 4000 series) were not designed to be TTL compatible and generally have active-high inputs. The 74C/74HC etc CMOS parts derived from similar TTL functions tend to have active-low inputs. Remembering this difference can serve you even today- if you need a flip-flip with active high clear input you are more likely to find it in a 4xxx or 74HC4xxx part than in a 74HCxx part such as 74HC74.
In some technologies, it's easier to make a N channel FET with more current capability than a P channel FET.
The 8086 pre-dates (fast, low voltage) CMOS logic. It's built in NMOS technology, with NMOS transistors (which, like NPN transistors, make good pull-downs) and ... I'm not sure what it used as pull-ups, possibly more NMOS transistors which, in that role, make slow, low current pull-ups.
So, analogous to contemporary TTL logic - which was used for much of the support logic, it was designed to the limitations of the available transistors, which meant critical signals were pulled down, and active low logic was easier to use for them.