What % of traffic is network overhead on top of HTTP/S requests

You have zero knowledge about the layers below HTTP. You can't even assume the HTTP request will be delivered over TCP/IP. Even if it is, you have zero knowledge about the overhead added by the network layer. Or what the reliability of the route will be and what overhead will be due to dropped/resent packets.

Update: Based on your comment, here are some back-of-the-napkin estimates:

The maximum segment size (which does not include the TCP or IP headers) is typically negotiated between the layers to the size of the MTU minus the headers size. For Ethernet MTU is usually configured at 1500 bytes. The TCP header is 160 bits, or 20 bytes. The fixed part of the IPv4 header is 160 bits, or 20 bytes as well. The fixed part of the IPv6 header is 320 bits, or 40 bytes. Thus:

• for HTTP over TCP/IPv4

overhead = TCP + IP = 40 bytes

payload = 1500 - 40 = 1460 bytes

overhead % = 2.7% (40 * 100 / 1460)

• for HTTP over TCP/IPv6

overhead = TCP + IP = 60 bytes

payload = 1500 - 60 = 1440 bytes

overhead % = 4.2% (60 * 100 / 1440)

Here are the assumptions:

• Amazon counts the NIC payload without the Ethernet headers, not the whole NIC packet
• your HTTP responses are fully utilizing the TCP/IP packet - your typical page size + HTTP headers results in one or more full TCP/IP packets and one with more than 50% used payload
• you set explicit expiration date on cached content to minimize 302 response
• you avoid redirects or your URLs are long enough to fill the payload

With 100Mbit/s Ethernet, a large file transfers at 94.1Mbit/s.

That's 6% overhead. If you also count the TCP ACKs flowing in the opposite direction, it's closer to 9%. For gigabit Ethernet the overhead (in percent) remains the same. Assumptions: TCP/IPv4 and file size >100kB. (At this size we can neglect the initial HTTP and TCP setup.)

When comparing download rates, beware the factor 8 from bits to bytes. I guess nobody will charge you for Ethernet preamble or interframe gap, but "payload" shouldn't be taken literally.

Calculation: payload / overall
payload = 1500 - 20 - 32 (Ethernet_MTU - IPv4 - TCP)
overall = 8 + 14 + 1500 + 4 + 12 (Preamble + Ethernet_header + Ethernet_MTU + CRC + Interframe_gap)

Because Ethernet is always full-duplex these days, the occasional TCP ACK flowing the other way does not change the transfer rate. If you add one ACK for every two data frames to the overhead (that's what I observed in Wireshark), you get 8.5% total overhead. And while the MTU size is usually 1500 bytes, it can be smaller in some networks, or much larger if every piece of equipment in the path is configured for it.