Q:

Maximize Arduino serial output

There are a multiple questions intersperced throughout this post. Kindly read carefully and answer the pieces you can for an upvote.
Use Case
Create a time series csv file of pressure reads. These reads need to be of maximum frequency, but I only need it to last less than 10 seconds or so.
Material

Arduino Uno Clone (unalterable)
serial over USB 2.0 (alterable)
pyserial (Python 3)
SSD

Problem
Identify and fix the bottleneck preventing frequency from maximum reads/s.
The code
Arduino
void setup() {
  //speed over the usb serial interface
  Serial.begin(450000);
}

void loop() {
    //get sensor data: read the input on analog pin 0:
    int sensorValue = analogRead(A0); //returns value 0-1023
    //send over serial
    Serial.println(sensorValue);
}

Python
...
ser.serial()
ser.baudrate=450000
ser.open()
while True:
    try:
        serialData = float(ser.readline().decode().replace('\n', ''))
    except(ValueError):
        continue #I ran the code without this try/except and got 100 reads higher than when this is included in here. Caused no change in order of magnitude.
    now = datetime.datetime.now()
    serialData2 = serialData * voltageOffset #a couple of these happen
    outfile.write(now+', '+str(serialData2)+'\n')
except(KeyboardInterrupt):
    ...
    os.sys.exit(0)

Bandwidths
Some calculations to figure out where the bottleneck resides.
100 µs to read analog pin
1 packet = 10 bits
serial wire speed: 450000/10 = 45000 packets / second
Arduino samples-into-serial/second rate:
100 µs * 1s/(10^-6 µs) = 1 x 10^4 samples / second
arduino silicon adapter: cheap cloned chip. Assumed to be underperforming as it is not an arduino-licensed product, but this is deemed acceptable loss as will be explained later. Assumed to have unlimited bandwidth, though this obviously will increase performance by an order of magnitude.
USB 2.0
packets/s = 480 mb/s * 10^6 b/mb * 1 packet/10 b = 48 x 10^6 packets / second
PC file write speed
400 MB/s = 320 x 10^6 packets/s
PC python script
Unknown. Presumed to be infinite? the datetime.now() query is known to take ~ 8 µs. 
Conclusion: The bottleneck is the serial baud rate.
Adjusting the baud rate confirms this. I'm finding that el-cheapo arduino uno only supports a maximum 450000 baud rate, when the arduino actually supports 2-7 million from what I can gather online. That being said, something is still wrong. 
The linchpin question
Recall that at this baud rate, we should be seeing pressure reads generated at a theoretical 45000 packets / second.
baudrate-data length-no parity-1 end bit present: 450000-8-n-1
Let sensorValue = 1023. Data looks like this going into println: 1023\n
When I run serial.println(sensorValue) in the arduino, how many packets are sent over the wire to the PC? How is this serialized into packets? How many packets will be sent over the wire? Which one of the following is true:
A. 16 bit int requires 2 packets to be transmitted. The \n exacts 2 more. Total = 4
B. sensorValue is converted to a string and sent over the wire with separate packets: 1 for 1, 1 for 0, 1 for 2, 1 for 3, 1 for \, and 1 for n. Total = 6
C. Other (help me optimize??)
Tests
I'm not seeing results as I would expect.
Let baud = 450000.
Changing value in the arduino code serial.println(value) produces different reads/second in my txt file:
value = readPin\n >> 5417 pressure reads/s
value = 1 >> 10201 reads/s
value = 1023 >> 4279 reads/s
This tells me a couple things. There is overhead when reading the analogPin's value. I can also conclude the arduino also only sends over multiple packets of data when it has to (1 > 8 bit of data, 1023 > 16 bit of data). Wait, what? That doesn't make very much sense. How does that happen?
Adjusting the baud rate causes the number of reads to change until the arduino maxes out. But lets find the throughput I should be expecting.
450000 b/s * (1 packet / 10b) * [1 read / (4 or 6 packets)] = 11250 reads OR 7500 theoretical reads/s
Actual = 5417 reads/s
Where did 2,000 reads disappear to?

A:

You are right that the bottleneck is the serial port (in the sense that you transmit data inefficiently). However, most of your other assumptions are wrong. 
What does baud rate stand for
Baud rate is the number of distinct symbol changes per second. Here it would be equivalent for bits per second. The baud rate would include all bits transmitted, not only data (start, stop, parity).
How 10-bit value is transmitted
As you have 8n1 transmission, you cannot send exactly 10 bits of data. It must be a multiple of 8, so 8, 16, 24, etc. 10 bits will be sent in two 8 bit parts, just like they are stored in Arduino memory as an int.
How does println() work
println() converts numbers to string. You can specify base for this conversion (DEC, BIN, HEX, OCT) - default is DEC, so 1023 will be transmitted as 4 bytes + \n which is a single byte (ASCII 10) and \r (ASCII 13) which is also a single byte. Total of 6 bytes. 1 will require 3 bytes - 1 for data and 2 for newline and carriage return.
How to make it faster

Without changing almost anything - use println(val, HEX) - you will need 5 bytes maximum for numbers greater than 255.
Using Serial.write() - this function puts raw binary data into serial, so if you want to send a value 10, it will send just one byte of value 10. However, when you want to send 10-bit variable, things get complicated - you need 2 bytes for that, and PC needs to know which is the first part and which is second to stitch it back together. You will need to come up with some form of simple transmission protocol to handle this, so probably some start/stop character.
Do you need 10-bits resolution? If you could get on with 8-bits you could transmit just plain data, without any additional characters. Also, as suggested in the comments, you could use faster ADC clock then.
Also suggested by KIIV - you could use ADC conversion interrupt that would trigger instantly when the measurement is completed.

If you still go with 10-bit, I would suggest 3-byte frame: 1 start byte, and 2 bytes of data. This way you could theoretically reach  ~18 000 frames per second, which is above the maximum analogRead() frequency.

