The more important question you need to be asking: What do my batteries do with regards to charge/discharge?
From voltages and their capacity, I am assuming you use Li-Ion.
The comment Will made on your question is especially valid for Li-Ion, since they can cause huge currents to flow when they aren't of equal charge, which happens when you replace one. But certainly still valid for many other types.
But, what I am mainly getting at: Li-Ion batteries will discharge from full to empty in pretty much a fixed time at a reasonably steady rate, which depends on their quality. High end cells from a reputable source will last between 1 and 2 years on a full charge. Generic brand batteries from a decent source will last about a year on a full charge. eBay/Ali batteries anywhere between a few weeks and a year.
And that's with no load added to them.
At 2.4Ah with 1 year of self discharge time, that's 365.25days * 24hours = 8766 hours on average. Which makes for a self discharge current of approximately 0.273mA. That means your MCU will not be contributing to the discharge time at all.
Of course it's still "only" 0.5-ish mA for two cells, so 3mA in a capacitor would impact the time. Effectively making it less than 3 months if it does follow the worst case limit (which the 3mA is). But if you already have a double battery with diodes, what's the point of a capacitor higher than a couple of μF?
What is much more common for your problem is using prime Litium batteries, like the coin cells "CR2032" and "CR2016", which have a self discharge time of 10years or more. At about 240mAh energy content in a larger cell, over three years your 8μA MCU will eat away 8μA * 8766hours * 3 = 210mAh, which means the cell will be about empty. That's $2.50 per 3 years of battery costs, but with a very good reliability of it working for long stretches of time.
It is then usually set up as:
simulate this circuit – Schematic created using CircuitLab
You don't need a switch, because the diodes will take care of that. When the main power is larger than the battery voltage, 3.3V for example, it will supply the MCU, when it falls away, the batteries will share the load through the diodes over time.
I selected BAT54, but they may not be the best. It's best to use a diode with a reverse leakage current of (much) less than half a μA, while also giving a reasonably low forward voltage at 2 times your expected MCU average drain current. If the reverse leakage current at 3V (we don't care about 50V, because we only have 3V batteries) is a few micro amp, that will contribute noticably, because D3 will let the batteries leak that amount back into the power supply.
As long as you always change the batteries one by one, the MCU will always have a battery supplying it and the two caps will easily cover small transients.